1 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Daksh Quant Question Bank "May Edition" By Shantanu Shukla Direction (1-5): Pie Chart (I) shows percentage distribution of total number of students in five different schools and pie chart (II) shows percentage distribution of total difference between number of boys and girls in these five schools. Read the data carefully and answer the questions. Q1. When one student from each A & C chosen as class monitor, then find the probability of both class monitors being boys (boys > girls, for both the schools)? (a) (c) 84 (d) (e) A 20% B 15% D 25% C 2 (a) 3 8 15 4 3 4 (e) 15 E Total difference between boys and girls = 50 Q3. Number of boys in school F are 5 less than that of in E and number of girls in F are same as E. Find probability of selecting 1 boy & 1 girl from school – F. (Boys > girls, in school E)? 40 (a) 87 (b) A 20% (c) 14 29 43 87 2 (d) 3 (e) B 20% D 20% A 2 56 (d) 5 D E 30% 72 25 (c) 7 C 22.5% B 25 Q2. If five students from each B & D are chosen for representing their respective school in a debate competition, then find the maximum possible probability of girls in remaining students? (b) A 72 5 (b) 9 25 Total students = 200 E 17.5% 35 C D E www.bankersadda.com 29 Q4. Ratio of number of boys to girls left school A is 1 : 2 and when two students chosen from remaining students of school 11 A, then probability of both being boys or both being girl is 18 . Find difference between number of boys and girls who left the school A (Boys > girls, in school A)? (a) 6 (b) Can’t be determined (c) 1 (d) 4 (e) 2 C 10% B 16 | www.sscadda.com | www.careerpower.in | Adda247 App Q5. Ritu invested her total saving in three different FD schemes A, B and C in the ratio of 5 : 4 : 6 on CI for two years at the rate of 10%, 15% and 20% respectively.If interest is calculated annually and interest from scheme B is Rs. 744 more than interest from scheme A then, find difference between interest received from scheme C and scheme B by Ritu? (a) Rs. 4185 (b) Rs. 4175 (c) Rs. 3840 (d) Rs. 4580 (e) Rs. 3250 Direction (6–10): Bar graph given below shows percentage of students passed out of total students in four different schools and percentage of girls passed out of total passed students in these four different schools in annual exam. Read the data carefully and answer the questions. 70 60 50 Q7. Total boys failed from school Q is 60% of total students failed from that school and total boys passed from same school is 1440. Total failed boys from R is 192 more than total female passed from same school. Find difference between total girls failed from Q & R, if boys failed from R is 42% less than total students failed from R. (a) 272 (b) 242 (c) 252 (d) 240 (e) 262 Q8. If total number of girls passed from S is 1125 and total 1 failed boys are 133 % more than that of total failed girls from 3 same school, then find difference between number of failed boys and failed girls from school S? (a) 1100 (b) 1000 (c) 1110 (d) 900 (e) 1200 Q9. If difference between passed boys and passed girls from P is 600 and total boys passed from S is 1350 , then find total failed students from S is what percent more than total students participated from school P ? (a) 15% (b) 10% (c) 12% (d) 5% (e) 20% 40 30 20 10 0 P Q % of students passed R S % of passed girls Q6. Total boys passed from P is 900 and total girls failed from Q is 640. If total girls failed from Q is 36% less than of total students failed from Q, then find ratio of total students participated in exam from Q to that of from P? (a) 5 : 9 (b) 5 : 7 (c) 3 : 5 (d) 5 : 6 (e) 4 : 7 Q10. Total students participated in exam from school R is 50% more than that of from P and difference between passed boys from both the school is 2220, then find average number of girls passed from both the schools? (a) 1100 (b) 1000 (c) 1390 (d) 900 (e) 1200 Directions (11-15): Line graph shows usual average speed of five cars A, B, C, D and E. Speed is given in meter/minute. Answer the question according to given data. 2000 1800 Speed (meters/minutes) 1750 1500 1500 1200 1250 1000 750 500 250 0 A 3 www.bankersadda.com | 900 750 www.sscadda.com | B C Cars www.careerpower.in D | E Adda247 App Q11. Car D started from Lucknow to Delhi at its usual speed for first half distance, but after that due to some problem in engine car travel at 4/6 of its usual speed. If car completed whole journey in 10 hr, then find the total distance between Lucknow and Delhi? (a) 440 km (b) 432 km (c) 442 km (d) 450 km (e) 452 km Q12. Car C starts from Pune and at the same time Car A starts from Mumbai towards each other, and at the time both meet one car has traveled 180 km more than other car. Find the distance between Pune and Mumbai? (a) 540 km (b) 520 km (c) 500 km (d) 520 km (e) 640 km Direction (16-17): In the following wrong number series a term is wrong as per pattern followed in the series. You have to find that wrong term and term which should replace that wrong term is your answer. (You have to mark correct term as per series not the wrong term mentioned in series) Q16. 32, 48, 80, 112, 176, 208, 272 (a) 36 (b) 116 (c) 206 (d) 276 (e) Series is right no need to replace any number Q17. 12, 70, 348, 1390, 4168, 8332, 8332 (a) None of the given option will replace the wrong number (b) 8334 (c) 4172 (d) 1392 (e) 350 Q13. Rajeev go to his village from the city by car B and return by car C. If his total travelling time is of 11 hours, then find the distance between city and his village? (a) 500 km (b) 510 km (c) 520 km (d) 540 km (e) 1080 km Direction (18-20): What will come in the place of (A), (B) & (C) in second (II) number series. (II) number series follows the same pattern as pattern of first (I) number series: Q14. Car E travels at its usual speed between city X and Y and take 480 minutes to complete total distance. But at the time of returning car E decreases its speed by 12 km/hr. Then find time taken by car E (in minutes) returning from city Y to X? (a) 526 minutes (b) 530 minutes (c) 576 minutes (d) 550 minutes (e) 612 minutes (c) 18.4, 34.4, 52 (d) 18.4, 34.4, 56 (e) 19.4, 34.4, 52 Q15. The distance between Delhi and Gorakhpur is 762 km. Car E starts at 4 pm from Delhi towards Gorakhpur at a given speed. Another car C starts at 3.20 pm from Gorakhpur towards Delhi at a given speed. How far from Delhi both cars meet and at what time? (a) 8:20 pm, 312 km (b) 7:20 pm, 290 km (c) 8:10 pm, 390 km (d) 6:20 pm, 350 km (e) 9.20 pm, 480 km 4 www.bankersadda.com | Q18. Series I – 2.4, 4, 7.2, 12, 18.4, 26.4, 36 Series II – (A), 20, 23.2, 28, (B), 42.4, (C) (a) 18.4, 34.4, 54 (b) 18.4, 36, 52 Q19. Series I – 123, 171, 227, 291, 363, 443, 531 Series II – (A), (B), 402, 486, (C), 678, 786 (a) 258, 326, 580 (b) 258, 326, 576 (c) 256, 326, 578 (d) 258, 324, 578 (e) 258, 326, 578 Q20. Series I – 20, 1300, 2100, 2660, 3100, 3480, 3830 Series II – (A), 1600, 2400, (B), 3400, 3780, (C) (a) 320, 2960, 4130 (b) 280, 2840, 4250 (c) 320, 2960, 4180 (d) 280, 2840, 4200 (e) 320, 2960, 4050 www.sscadda.com | www.careerpower.in | Adda247 App Directions (21-25) : An exam consists of 100 questions among 3 sections i.e. A, B and C. Section C have 32 questions and section A and B have equal number of questions. For each correct answer a student is awarded 3 marks and 1 mark is deducted for every wrong answer. Each section has 3 questions for which there is no negative marking on wrong attempt and 0.50 marks are deducted for every unattempted question. Amar and Prem appeared for same exam, Amar attempted total 75 questions, while Prem attempted 80% of total questions attempted by Amar and 90% of his questions were correct. Both of them attempted all the question which do not have negative marking. Q21. What is the smallest range of marks obtained by Amar, if 80% of his attempted questions were correct? (a) 165-174 (b) 165-174 (c) 152.5-161.5 (d) 165-180 (e) 155-165 Q22. If only 3 non-negative marking questions out of 9 were wrong for Prem. Find the score of Prem. (a) 122 (b) 102 (c) 159 (d) 139 (e) None of these Q23. Prem attempted all the questions of section A, which were all correct and 50% of questions section C in which 6 were incorrect carrying negative marking. Find the difference between his score from section C and section B. (a) 03 (b) 02 (c) 14 (d) 06 (e) None of these Q24. All the non-negative marking questions were correct for Amar and only non-negative marking questions were wrong for Prem. What is the difference between their score, If 80% of questions attempted by Amar were correct. (a) 10.5 (b) 16.5 (c) 54.5 (d) 36.5 (e) 17.5 Q25. If Amar scored 108.5 and 5 of his non-negative marking questions were correct. What is the total number of his correct questions? (a) 42 (b) 53 (c) 47 (d) 49 (e) None of these 5 www.bankersadda.com | Directions (26-30): Study the graph and table given below and answer the following questions. The line graph shows the listed price per kg of various items in a wholesale store. 45 40 35 30 25 20 15 10 5 0 The table given below shows the amount of items bought by a retailer from the wholesale store. The table also shows the discount % offered by the wholesaler on the list price and total cost incurred by the retailer. Quantity Discount Total Items (in kgs) (in %) (in Rs.) Rice 20 10 Wheat 30 675 Sugar 15 240 Pulses 18 30 Cashew 40 900 Almond 25 15 Q26. Calculate the profit earned by retailer on selling 20 kgs of wheat purchased by him to a customer at a discount of 5% on the listed price ? (a) Rs. 25 (b) Rs. 45 (c) Rs. 75 (d) Rs. 50 (e) None of these Q27. The retailer sold all the cashew bought by him to a customer at a price 25% more than the listed price. Calculate his overall profit percent. (a) 33.33% (b) 66.66% (c) 55.55% (d) 42.64% (e) 77.77% Q28. If 50% of the rice bought by the retailer got spoiled, then at what price/kg must he sell the remaining amount of rice to be at a situation of no loss-no gain ? (a) Rs. 40 (b) Rs. 19 (c) Rs. 27 (d) Rs. 22 (e) None of these www.sscadda.com | www.careerpower.in | Adda247 App Q29. The retailer sold all the pulses he bought at a price that is 30% more than the listed price and offered 2 kgs of Almond free with it. Find overall profit% of the retailer in this bargain? (approximate) (a) 50% (b) 40% (c) 35% (d) 61% (e) 45% Q30. The retailer mixed 6 kgs. of impurity (free of cost) with all the sugar he had and sold the mixture at a discount which is 25% less than that discount (in percentage) offered by the wholesaler. Find the profit % on the sale of all of the amount of this mixture? (a) 52.50% (b) 46.15% (c) 48.75% (d) 57.50% (e) None of these Direction (31–35): Table given below shows data regarding number of people applied for loan under ‘PM Mudra Yojna’ from five different villages. Read the data carefully and answer the questions. Ratio of Percentag Percentag femal e of people e of male e who Numbe who get who get do not r of loan out of Village loan out of get people total s total loan applied number of people to for loan people who get femal applied for loan e who loan get loan P 7200 66 3 % 2 65% 10 : 21 Q 8000 60% 75% 3:5 R 8800 9 81 11 % 82% 4:9 S T 10000 9600 72 % 68.75% 76% 80% 16 : 27 8 : 11 Q31. What is the difference between number of males applied for loan from village P and village T? (a) 2600 (b) 2200 (c) 2400 (d) 3000 (e) 2000 6 www.bankersadda.com | Q32. Total females who do not get loan from village S is what percent more or less than total females who do not get loan from village Q? 2 (a) 40 9 % 2 (b) 44 % 9 2 (c) 46 % 3 2 (d) 48 3 % 2 (e) 42 9 % Q33. What is the ratio of total males who do not get loan from village Q to total males who do not get loan from village S? (a) 113:111 (b) 115: 111 (c) 64: 111 (d) 155: 111 (e) 111 : 115 Q34.How many females applied for loan from the village where number of males who get loan are second highest among all villages? (a) 2772 (b) 2726 (c) 2752 (d) 2742 (e) 2732 Q35. What is the average of number of males who do not get loan from village R and number of females who get loan from village R? (a) 1120 (b) 1140 (c) 1260 (d) 1200 (e) 1160 Direction (36-40): Find the wrong number in the following number series given below : Q36. 5, 86, 174, 276, 399, 558, 736 (a) 276 (b) 736 (c) 558 (d) 86 (e) 399 Q37. 9, 4.5, 6.5, 14, 57, 457, 7313 (a) 4.5 (b) 57 (c) 457 (d) 9 (e) 7313 www.sscadda.com | www.careerpower.in | Adda247 App Q38. 1728, 998, 1511, 1167, 1384, 1260, 1323 (a) 998 (b) 1511 (c) 1323 (d) 1167 (e) 1260 Directions (44-45): Answer these questions based on the information given below. 6 men complete a piece of work in 12 days. 8 women can complete the same piece of work in 18 days. Whereas 18 children can complete the piece of work in 10 days. 4 men, 12 women and 20 children work together for 2 days, and then only 36 men were to complete the remaining work in 𝑥 day. Q39. 2.5, 60, 720, 4320, 12960, 19480, 14580 (a) 720 (b) 4320 (c) 12960 (d)19480 (e) 14580 Q44. A can do a piece of work in 10𝑥 days and B can do the same work in 20𝑥 days. They do the work on alternative days starting from A then in how many days A and B can complete the work? (a) 11 days (b) 12 days (c) 13 days (d) 14 days (e) None of these Q40. 11.5, 34, 58, 85, 116.5, 154, 200 (a) 200 (b) 85 (c) 116.5 (d) 154 (e) 34 Q41. Working alone, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts, they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together? 1 (a) 5 days 3 1 (b) 5 7 days 5 (c) 5 9 days 5 (d) 5 11 days (e) None of these Q42. Bag A contains ‘P’ green and 18 yellow balls while bag B contains ‘(P + 2)’ green balls and 22 more number of yellow balls than that of in bag A. Probability of selecting a green ball 1 from bag A is 12 more than probability of selecting a green ball from bag B. Find total number of balls in bag B. (P<50) (a) 48 (b) 54 (c) 60 (d) 84 (e) 66 Q45. 56𝑥 soldiers can complete a piece of work in 24 days. In how many days can 42 soldiers complete the same piece of work? (a) 32 days (b) 24 days (c) 16 days (d) 48 days (e) None of these Directions (46-47): Three friends P, Q and R share an apartment and share the rent equally. The monthly income of R is 25% less than that of Q and Rs.8000 less than that of P. Monthly expenditure of Q on food is Rs.1000 more than that of P and is Rs.1000 less than that of R. After meeting the expenses on rent and food, they save amounts in the ratio 6 : 7 : 4. 1 Q46. If Q saves 62 2 % of his total monthly income, then how much percent does R save out of his monthly income? 13 (a) 47 21 % 12 (b) 48 21 % 5 (c) 45 21 % 11 (d) 49 % 21 (e) Cannot be determined Q43. Ratio of speed of boat P and Q in still water is 2 : 3. If time taken by boat ‘Q’ to cover a certain distance in upstream is equal to that of ‘P’ to cover 60% of that distance in upstream and difference of the distance covered by ‘Q’ and ‘P’ in 5hr in downstream is 60km then find ratio of speed of boat ‘P’ in downstream to ‘Q’ in upstream? (a) 3 : 2 (b) 1 : 2 (c) 1 : 1 (d) 4 : 5 (e) 5 : 4 7 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q47. If the total amount spent by all the three on food is Rs.27000 and the monthly income of Q is Rs.6000 more than that of P, then what is the monthly rent of the apartment? (a) Rs.48000 (b) Rs.30000 (c) Rs.24000 (d) Rs.36000 (e) Cannot be determined Q48. Selling price of two articles is same and shopkeeper claim to get 20% profit on each article but mistakenly he calculates one profit on selling price. If difference of profit he gets on articles is 8 Rs. then find the selling price? (a) Rs.200 (b) Rs.180 (c) Rs.160 (d) Rs.220 (e) Rs.240 Q52. If the number of girls in sections A is same as the number of boys in section C, then what is the ratio of number of boys in section A to the number of boys in section C? (a) 2 : 3 (b) 3 : 4 (c) 3 : 2 (d) 4 : 3 (e) None of these Directions (53-55): Given below are two pie charts. Pie chart (1) shows the percentage distribution of milk in five vessels out of the total milk in these five vessels. Pie chart (2) shows the percentage distribution of water in same five vessels out of total quantity of water in these five vessels. (1) (2) E 20% Q49. Rs.15000 invested in two schemes each. First scheme offers R% interest on C.I. and second scheme offer R% more of what he gets in first at S.I. If after 2-year difference between interests earned in both scheme is Rs. 600 then find the value of R. (a) 24% (b) 10% (c) 15% (d) 20% (e) None of these A 15% B 20% D 10% C 35% Q50. Sunny and Satish shoot each other 4 and 3 out of every 6 shots respectively. They try to shoot each other in sequences start with Sunny. Find the probability that Satish killed? 1 (a) 5 D 5% E 15% A 10% B 25% 2 (b)5 7 (c)10 C 45% 3 (d)5 4 (e)5 Directions (51-52): There are three sections A, B and C in a class. Every section has some boy and some girl students in it. Probability of a girl being selected when one student is 2 4 selected randomly from section A is 5, that from section B is 9 5 and that from section C is . 9 Q51. If the ratio of total number of students in sections A, B and C is 10 : 12 : 9, then what is the probability of a girl being selected when one student is selected randomly from the students from all the three sections together? 14 (a) 31 (b) (c) Note: Ratio of total milk to total water in these five containers is 2 : 1. Q53. A shopkeeper pours the mixture of vessel A and B into another vessel F. Vessel F contains water only which is equal to 25% of water of vessel B. If shopkeeper professes to sell the whole mixture at the cost price of pure milk and cost price for shopkeeper is due to milk only, then find the percentage profit of shopkeeper in selling whole mixture. 13 (a) 5814% 13 11 (b) 315% 23 13 (c) 54 % 31 43 (d) 53 % 13 15 13 (d) 93 (e) Cannot be determined 8 www.bankersadda.com 14 20 (e) 5521% | www.sscadda.com | www.careerpower.in | Adda247 App Q54. Mixture of vessel A and C are mixed into another vessel M. If 62 liters of the mixture M is taken out and replaced with 17 L of water, the ratio of milk to water in M becomes 6 : 5. Find the quantity of milk in vessel B. (a) 60 L (b) 20 L (c) 40 L (d) 45 L (e) 50 L Q55. All the contents of mixture from all vessels except C is poured into bigger vessel and from vessel C, only 115 liters of mixture is taken out and poured into bigger vessel, then ratio of milk and water in bigger vessel becomes 9 : 4. Find the total quantity of water in all five vessels. (a) 550 L (b) 500 L (c) 600 L (d) 650 L (e) 700 L Directions (56-60): The following questions are accompanied by three statements (A), (B), and (C). You have to determine which statement(s) is/are sufficient necessary to answer the questions Q56. A bag contains some balls of red, black and blue colors. 3 balls are drawn randomly from it. What is the probability that the balls drawn are of three different colors? A. Ratio of black and blue colored balls is 4 : 5. B. Total number of red and black balls is 5 more than that of blue colored balls. C. Total number of balls in that bag is 35 and probability of 3 selecting blue colored balls is 7. (a) A and either B or C (b) Any two of them (c) Only A and C together (d) Question can't be answered even after using all the information (e) All statements are required Q57. Find area of a rectangle if its perimeter is equal to the perimeter of a square. A. Length and breadth of the rectangle are in the ratio of 4 :3 and side of a square is half of the sum of length and breadth of that rectangle. B. Sum of the lengths of diagonals of the rectangle is 4m less than thrice of the side of square. C. Area of a square is 784 m2 (a) Only A and B together (b) Only A and C together (c) All the three together (d) Any two of the three together (e) Either A and B together or A and C together 9 www.bankersadda.com | Q58. What is the value of a two digit number? A. The sum of the number and its square is 25 times the number itself. B. The number obtained after interchanging the digits is greater than the original number by 18. C. The ratio of the value of the number and the sum of the digits of that number is 4 : 1. (a) Either A alone or B and C together (b) All the three together (c) Any one of the three (d) All the three together are not sufficient (e) Either A or B alone Q59. What is the selling price of an article if the marked price of that article is 60% above the cost price of that article? A. The article is sold at 12% profit and there are two 1 successive discounts of 20% and 12 2 % on marked price. B. Difference of marked price and the cost price of that article is Rs 480. C. The marked price is Rs.384 more than the selling price of that article. (a) Only B and C together (b) Any two of three are sufficient (c) Either A and B together or A and C together (d) Only A and C together (e) All the three together Q60. Three friends A, B and C invested in a scheme. Find the profit share of A? A. Investment of A, B and C is in the ratio 9: 12: 10 and share of C in profit is Rs.1800 B. A and B invested Rs.8100 and Rs.10800 for 10 months respectively and profit share of B is 100% of the profit share of C. C. A invested Rs 900 more than that of C and investment period of B is 10 months. (a) A and B together only (b) Either A and B or A and C (c) Any two of them (d) Either B alone or A and C together (e) Either A and C or B and C Directions (61-65): The following questions are accompanied by three statements A, B and C. You have to determine which statement(s) is/are necessary/sufficient to answer the question. Q61. A shopkeeper sells articles at a certain profit. Find out the amount of profit. A. Ratio of the selling price to the cost price of the articles is 5 : 4. B. If the cost price increases by Rs.500, and selling price 8 remains the same, the profit percentage is decrease by 139 percentage points. C. If the marked price is kept at Rs.1000 above the cost price and a discount of 15% is given, then the profit percentage is 3 decreased by 18 4 percentage points. (a) Only A and B together (b) A and either B or C (c) Only A and C together (d) All statements are required (e) None of these www.sscadda.com | www.careerpower.in | Adda247 App Q62. Rinku borrowed an amount of Rs.5000 each from Milan and Rahul. What is the rate of interest? A. Rinku returned the amount of Rs.5400 after due date to Milan. B. Rinku returned Rs.5900 to Rahul after due date. C. Rinku returned the money to Milan by SI, whereas to Rahul by compound interest. (a) Only A and B together are sufficient (b) Only B and C together are sufficient (c) A, B and C together are necessary (d) Either A and B together or B and C together are sufficient (e) A, B and C even together are not sufficient Q63. What is the speed of boat in still water? A. The boat can cover 45 km downstream distance in 3 hours. B. Speed of the stream is one-fourth the speed of boat in still water. C. The boat can cover 36 km upstream distance in 4 hours. (a) Only (A) and (C) together (b) All the three together (c) Only (A) and (B) together (d) Questions can’t be answered even after using all the information (e) Any two of the three together Q64. A train crosses another train in 10 sec. Find out the lengths of the trains. A. Ratio between the lengths the of second and first train is 4 : 5. B. Ratio between the speed of first and second trains is 1 : 2. C. The speed of first train is 36 km/hr. (a) Only A and B together (b) Only B and C together (c) Only A and C together (d) Questions can’t be answered even after using all the information (e) None of these Q65. Find the height of an equilateral triangle. A. Perimeter of the triangle is equal to the perimeter of the rectangle whose length and breadth are in the ratio of 5 :3. B. Perimeter of the square is 48 m, which is twice the perimeter of the triangle. C. Area of the triangle is 16√3m2. (a) Any two of them (b) Any of them (c) Only C (d) Either B or C alone (e) A and either B or C 10 www.bankersadda.com | Directions (66-70): In the given questions, two quantities are given, one as Quantity I and another as Quantity II. You have to determine relationship between two quantities and choose the appropriate option Q66. In a two-digit number, digit at unit place exceeds, the digit in its tens place by 2 and the product of the required number with the sum of its digit is equal to 144. Quantity I: Value of two digit number Quantity II: 26 (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q67. Quantity I : Days after which A and B meet. A and B set out to meet each other from two places 165 km apart. A travels 15 km the first day, 14 km second day, 13 km the third day and so on, B travels 10 km the first, 12 km the second day, 14 km the third day and so on. Quantity II: Number of days required to complete the whole work if A, B and C can complete a piece of work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q68. Quantity I: Present age of Randy, if 10 years are subtracted from the present age of Randy, then you would get twelve times of the present age of his grandson Sandy and Sandy is 19 years younger to Sundar whose age is 24. Quantity II: Average age of the remaining persons in the group if average age of group of 14 persons is 27 years and 9 months. Two persons, each 42 years old, left the group. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q69. Quantity I: Percentage profit earned by the shopkeeper if at the time of selling and purchasing he uses weights 10% less and 20% more per kilogram respectively and professes to all goods at 5% profit. Quantity II: ‘x’ ; A book was sold for a certain sum and there was a loss of 20%. Had it been sold for Rs 12 more, there would have been a gain of 30%. ‘x’ would be value of profit percent if the book were sold for Rs 4.8 more than what it was sold for. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation www.sscadda.com | www.careerpower.in | Adda247 App Q70. A group consist of 4 couples in which each of the 4 persons have one Quantity I : Number of ways in which they could be arranged in a straight line such that the men and women occupy alternate positions Quantity II: Eight times the number of ways in which they be seated around circular table such that men and women occupy alternate position. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Directions (71-74): The following questions are accompanied by two statements I and II. You have to determine which statements(s) is/are sufficient/necessary to answer the questions. Q71. Find the sum of present age of X & Z? I. Four years ago, average age of X, Y and Z was 16 years while age of X, Y and Z is in arithmetic progression. II. Average of present age of X and Z is same as present age of Y while Y is four years younger than Z. (a) Statement I alone is sufficient to answer the question but statement II alone is not sufficient to answer the question. (b) Statement II alone is sufficient to answer the question but statement I alone is not sufficient to answer the question. (c) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (d) Either statement I or statement II by itself is sufficient to answer the question. (e) Statements I and II taken together are not sufficient to answer the question. Q72. Is Rahul meet Prabhat at half of the distance on a circular track (48km) if they both started to move at same time in same direction? I. Speed of Prabhat is 24km/hr more than speed of Rahul. II. Prabhat’s speed is 200% more than that of Rahul (a) Statement I alone is sufficient to answer the question but statement II alone is not sufficient to answer the question. (b) Either statement I or statement II by itself is sufficient to answer the question. (c) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (d) Statement II alone is sufficient to answer the question but statement I alone is not sufficient to answer the question. (e) Statements I and II taken together are not sufficient to answer the question. 11 www.bankersadda.com | Q73. Find number of yellow balls in a box which contains 3 black balls and 7 red and yellow balls. 2 I. Probability of selecting two yellow balls from the box is 15 1 II. Probability of selecting one red and black ball is 5. (a) Statement I alone is sufficient to answer the question but statement II alone is not sufficient to answer the question. (b) Statement II alone is sufficient to answer the question but statement I alone is not sufficient to answer the question. (c) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (d) Either statement I or statement II by itself is sufficient to answer the question. (e) Statements I and II taken together are not sufficient to answer the question. Q74. Find number of pens sold by retailer if he earns total 37.5% profit on selling some pens and some pencils. I. Ratio between pen sold to pencil sold is 2 : 3 while ratio between cost price of pen to pencil is 3 : 2. II. On selling a pen and a pencil, he earns 25% and 50% profit respectively. (a) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (b) Statements I and II taken together are not sufficient to answer the question. (c) Statement II alone is sufficient to answer the question but statement I alone is not sufficient to answer the question. (d) Statement I alone is sufficient to answer the question but statement II alone is not sufficient to answer the question. (e) Either statement I or statement II by itself is sufficient to answer the question. Directions (75-76): In the given questions, two quantities are given, one as Quantity I and another as Quantity II. You have to determine relationship between two quantities and choose the appropriate option. Q75. There are three vessel A, B and C. Vessel A contains mixture of milk and water in ratio 3 : 1. Vessel B contains 20 litres of pure water while vessel C contains 30 litres of pure milk. Half of the content of vessel A is first poured into vessel B. Then content of vessel B is poured into vessel C and finally contents of vessel C is poured into vessel A. The final ratio of milk and water in vessel A is 9 : 4. Quantity I: Initial quantity of mixture in vessel A. Quantity II: 80 litres. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation www.sscadda.com | www.careerpower.in | Adda247 App Q76. 20 men can complete a work in 12 days. 5 women are as efficient as 3 men. 4 men and 10 women started working and they already worked for 8 days. Quantity I: Additional number of women required to complete the remaining work in 10 days. Quantity II: Additional number of men required to complete the remaining work in either 8 or less than 8 days. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q77. Quantity I — Time taken by A to complete a work alone if A can complete a work in 5 more days than B while A does the same work in 9 more days than C. If A and B can complete the whole work in same time as time taken by C alone to do the whole work. Quantity II — Time taken by 8 men and 14 women to reap 7 𝑝𝑎𝑟𝑡 of 360 hectare land by working 7 hrs per day if 6 men 12 5 and 10 women can reap 12 part of the land in 15 days by working 6 hrs per day. It is also given that work of 2 men is equal to that of 3 women. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q78. Quantity I — Difference between the speeds of P and Q if 2 places A and B are 60 km apart. P and Q start from A at same time & meet 1st time at a place 12 km from B & they reach A after immediate return from B. The speed of slower person is 48 km/hr. Quantity II — Average speed of train if a distance of 600 km is to be covered in 2 parts. In 1st phase 120 km is travelled by train and rest by car and it took total of 8 hrs, but if 200 km is covered by train and rest by car it takes 20 min more. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q79. Quantity I — Cost price of motor bike if a man promises to pay the price of the motorbike in 3 equal annual installments of 10,800 Rs. at the compound interest rate of 20% per annum. Quantity II — 240% of the value of each installment if a man borrowed a sum of Rs. 25220 from a bank and promise to pay the amount in 3 equal installments at the compound interest rate of 5% per annum. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation 12 www.bankersadda.com | Q80. Quantity I — Value of 𝑥, if ABCD is a rectangle and AB= 10 unit, AD= 6 unit. Quantity II — value of 𝑦, if volume of the cone is 16𝜋 unit3 Radius = 4 unit (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or No relation Q81. Sandeep bought some chilli powder containing five percent impurities in the form of raw dust. He then mixed pure chilli powder to 2 kg of that impure chilli in such a way that the proportion of impurity now became 4% of total mixture. At what percent should he mark-up the price of this mixture to have an overall profit of 30% ? (a) 25% (b) 35% (c) 20% (d) 32% (e) None of these Q82. In an examination, there are 100 questions. For each correct answer 4 marks will be awarded, for each wrong answer 2 mark will be deducted for each unattempted question 1 mark will be deducted. If a student scored 320 marks then what could be max number of questions attempted wrong by him? (a) 12 (b) 10 (c) 14 (d) 13 (e)None of these Q83. A and B can do a piece of work in 12 days. B and C together can do the same work in 16 days while B alone can do it in 24 days. Two of them who have same efficiency work on day one and then the third person does the work on day two. If this process goes on till the completion of the work, then find the total no. of days taken for the work to get finished ? 3 (a) 15 days 4 3 (b) 17 days 4 3 (c) 16 days 4 3 (d) 18 days 4 (e) None of these www.sscadda.com | www.careerpower.in | Adda247 App Directions (84-85): Abhishek bought some chairs and tables from a shopkeeper. The marked price of a chair and a table were in the ratio 5 : 8. The shopkeeper gave discounts of 20% and 25% on the chair & the table respectively. The ratio of number of chairs and tables bought by Abhishek is 6 : 5. Q84. If Abhishek sells each chair and table bought by him at discounts of 25% and 20% respectively after marking up the prices of both by 50% and gives one table free for every four 2 chairs bought by a customer and only 3 rd of the total chairs are sold in bunch of four chairs, then what is the net profit /loss % made by Abhishek after selling all of the items which he bought from the shopkeeper? 2 (a) 6 3 % 1 (b) 3 % 3 1 (c) 2 % 2 1 (d) 4 4 % (e) None of these Q85. If the marked price of a table set by the shopkeeper was Rs.300 more than that of a chair and the total expenditure made by Abhishek in purchasing the chairs and table from the shopkeeper was Rs.108000, then how many chairs were purchased by Abhishek? (a) 150 (b) 60 (c) 120 (d) 90 (e) None of these Directions (86-87): A team of miners planned to mine 1800 tons of ore during a certain number of days. Due to technical difficulties, in one-third of the planned number of days, the team was able to achieve an output of 20 tons of ore less per day than the planned output. To make up for this, the team overachieved for the rest of the days by 20 tons. The end result was that the team completed the task one day ahead of time. Q86. How many tons of ore did the team initially plan to mine per day? (a) 50 (b) 100 (c) 150 (d) 200 (e) 250 Q87. To complete the task two days before the planned date, how many more tons of ore should the team mine per day after mining the planned tons of ores till one-third of the planned no. of days? (a) 25 (b) 30 (c) 20 (d) 15 (e) None of these 13 www.bankersadda.com | Direction (88-89)- In a school there are a total of 50 students in class X, who are divided into three sections A, B and C. A and B have an equal number of students. All the students of the class wrote a test. The average marks obtained by the students of sections A and B together is 52.5. The average marks obtained by the students of sections A and C together is 60. The average marks obtained by the students of sections B and C together is 70. The average marks obtained by the students of sections A, B and C together is 60. Q88. How many students are there in section C? (a) 10 (b) 20 (c) 15 (d) Cannot be determined (e) None of these Q89. How many students are there in all of the three sections? (a) 40 (b) 30 (c) 45 (d) Cannot be determined (e) None of these Q90. Ratio of length and speed of two trains (smaller: longer) are 7: 8 and 4: 3 respectively. Time taken by both trains to cross each other while running in opposite directions is 6 12 7 𝑠𝑒𝑐 and time taken by smaller train to cross a platform having length 100 m less than the smaller train is 16 sec. Find time taken by longer train to cross that platform? 1 (a) 22 2 𝑠𝑒𝑐 1 (b) 23 3 𝑠𝑒𝑐 2 (c) 21 3 𝑠𝑒𝑐 (d) 20 sec (e) None of the given options Directions (91-93): The following questions are accompanied by two statements (I) and (II). You have to determine which statements(s) is/are sufficient/necessary to answer the questions. (a) Statement (I) alone is sufficient to answer the question but statement (II) alone is not sufficient to answer the questions. (b) Statement (II) alone is sufficient to answer the question but statement (I) alone is not sufficient to answer the question. (c) Both the statements taken together are necessary to answer the questions, but neither of the statements alone is sufficient to answer the question. (d) Either statement (I) or statement (II) by itself is sufficient to answer the question. (e) Statements (I) and (II) taken together are not sufficient to answer the question. www.sscadda.com | www.careerpower.in | Adda247 App Q91. What is volume of cone? I. Radius of cone is 3 cm less than side of square, whose area is 576 cm2. II. Height of cone is 7.5 cm more than radius of circle, which circumference is 66 cm. Q92. What is rate of interest? I. A man invested an amount for three years on simple interest and gets a total amount, which is 137.5% of invested amount. II. Amir invested Rs. 9600 on simple interest and gets a total amount of Rs.13200 after three years. Q93. The ratio between length of two trains is 9 : 8. What will be difference between lengths of both trains? I. Speed of larger train and smaller train is 72 km/hr and 90 68 km/hr respectively. Both trains cross each other in 𝑠𝑒𝑐 9 running in opposite direction. II. Speed of smaller train is 90 km/hr and it cross a pole in 6.4 sec. Directions (94-96): In the given questions, two quantities are given, one as ‘Quantity I’ and another as ‘Quantity II’. You have to determine relationship between two quantities and choose the appropriate option: Q94. Quantity I – A bag contains five red balls, six green balls, ‘a’ yellow balls & ‘b’ blue balls. Probability of drawing one 1 yellow ball is , while probability of drawing one blue ball is 6 2 . If two balls are drawn from bag without replacing, then find 9 probability that one of them is red and other is yellow. Quantity II – A bag contains dices only in three colors, eight green color dice, ‘x’ blue color dice and ‘y’ yellow color dice. 7 Probability of drawing one blue dice is 20 , while probability of 1 drawing one yellow dice is 4. If two dices drawn at random without replacement, then find probability that one of them is blue and the other is green. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation Q95. Quantity I – A cylindrical vessel with radius and height of 17.5 cm and 18 cm respectively is filled upto 80% of its capacity with milk. If total milk from cylindrical vessel transferred into 30 cuboidal vessels whose length and breadth is 7 cm & 3 cm respectively. Find height of each cuboidal vessel? Quantity II – Breadth of a rectangle is 18 cm and ratio between length of rectangle and side of square is 12 : 11. If perimeter of square is 4 cm more than perimeter of rectangle. Find side of square. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation 14 www.bankersadda.com | Q96. Quantity I – A boat takes double time in covering same distance in upstream as compared to downstream, if boat covers 96 km in downstream and 72 upstream in total 20 hours. Find time taken by boat to cover 240 km in downstream. Quantity II – Distance between point A and point B is 720 km. 1 A car covered 𝑟𝑑 𝑜𝑓 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 with its usual speed and 3 remaining with 20% increased speed, if car takes total 10 hours 40 minutes to cover total distance, then find in what time car will cover a distance of 1200 km with its usual speed. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I ≥ Quantity II (d) Quantity I ≤ Quantity II (e) Quantity I = Quantity II or no relation Q97. A bank offers SI of 10% on principal amount below Rs.5000 otherwise 20%. A man invested Rs. A for 3 years. What is the value of A? (A) If he had submitted Rs. 3000 more, he will get an interest of Rs. 900 more. (B) If he had submitted Rs. 4000 more, he will get an interest of Rs. 2400 more. (C) Value of A is multiple of 500. (a) Either A and B or B and C are sufficient to answer the question (b) Either A and B or A and C are sufficient to answer the question (c) Either A and C or B and C are sufficient to answer the question (d) A, B and C together are sufficient to answer the question (e) All of the statements together are not sufficient to answer the question. Q98. A man has two items A and B. What is the selling price of item B? (A) The amount obtained after selling 4 of item A and 1 of item B is Rs. 70. From this amount he either could buy 7 of item A or 1 of item A with 4 of item B. (B) Profit earned on selling 1 of item B is Rs. 3 and profit % earned on selling 1 of item A is 30 %. Selling price and cost price of item B are Rs. 5 greater than that of item A respectively. (C) Ratio of profit % earned on A and B is 3 : 2 and ratio of their cost price is 2 : 3 respectively. All profit % , cost price and selling price have integer value. (a) Either A and B or B and C are sufficient to answer the question (b) Either A and B or A and C are sufficient to answer the question (c) Either A and C or B and C are sufficient to answer the question (d) A, B and C together are sufficient to answer the question (e) Either only B or A and C together are sufficient to answer the question www.sscadda.com | www.careerpower.in | Adda247 App Q99. Is |a × b| ≥ 10? (A) a + b = –3 (B) a × b < 0 (C) a + 2.5b = 0 and both a & b are integers (a) Only statement ‘A’ alone is sufficient to answer the question (b) Only statement ‘B’ alone is sufficient to answer the question (c) Only statement ‘C’ alone is sufficient to answer the question (d) Any of two statements are sufficient to answer the question (e) All three together are sufficient to answer the question Q100. Three friends Sonu, Monu and Jonu live together. Age of two of them is same. What is the age of Monu? (A) Average age of all of them together is 32 years, 4 years lesser than age of Sonu. 1 (B) Age of Monu is 333% lesser than age of Jonu and age of Monu is 8 years lesser than the averge age of all three. (C) Average age of 2 of them is 2 years lesser than average age of all three together. Monu is youngest among them. (a) Either A or B alone is sufficient to answer the question (b) Either B or C alone is sufficient to answer the question (c) Either A or C alone is sufficient to answer the question (d) Any of the two statements are sufficient to answer the question (e) Either only A or B and C together is sufficient to answer the question Directions (101-105): Given below the table shows types of interest offered by five banks, principal amount, time of period and rate of interest. Some of the data is missing. Calculate that according to information given in questions. Bank ICICI SBI YES UCO IDBI Type of interest Compound Simple Compound Compound Simple Principle (Rs) –– –– 20000 –– 10000 Time (year) –– 4 3 2 –– Amount (Rs) –– 26250 –– 29160 – Rate 15% –– 10% –– 6% Q101. If the ratio of interest rate of IDBI to that of UCO is 3: 4, then find the difference between principle invested in UCO bank and amount obtained from IDBI, if time period for both banks is same? (a) 13800 Rs (b) 12800 Rs (c) 11800 Rs (d) 13600 Rs (e) 13900 Rs. Q102. If rate of interest offered by SBI and Yes bank is same. Then find principle invested in SBI is approximately what percent of amount obtained from YES bank? (a) 52% (b) 59% (c) 70% (d) 65% (e) 78% 15 www.bankersadda.com | Q103. What is amount of interest obtained from ICICI bank, if ratio of principle invested in ICICI bank to principal invested in Yes bank is 7 : 5 and time period is one year less for ICICI bank than time period of YES bank? (a) 9020 Rs (b) 9030 Rs (c) 8030 Rs (d) 7030 R (e) 9080 Rs. Q104. Principle invested in ICICI is 3000 more than principle invested in UCO bank and both invested for same period of time and UCO bank offered 8% rate of interest annually. If amount obtained from ICICI is Rs. 32870 more than interest obtained from UCO bank then find the principle invested in UCO bank and ICICI bank? (a) Rs 25000 & Rs 27000 (b) Rs 18000 & Rs 16000 (c) Rs 22000 & Rs 20000 (d) Rs 25000 & Rs 28000 (e) Rs 24000 & Rs. 28000 Q105. If ratio between rate of interest offered by SBI bank to IDBI bank is 5 : 3 and ratio between time period is 2 : 1 respectively, then find the sum of principle invested in SBI bank and amount obtained from IDBI bank? (a) 27850 Rs (b) 28850 Rs (c) 29950 Rs (d) 27950 Rs (e) 31950 Rs. Directions (106–110): The bar graph shows % of sweets sold by a famous shop out of the total sweets that he prepared. Answer the questions based on this information. Note: All the units of sweets are given in KG’s unless mentioned. Barfi Laddu's 100 90 80 70 60 50 40 30 20 10 0 www.sscadda.com Friday | Saturday Sunday www.careerpower.in | Monday Adda247 App Q106. Shopkeeper prepared 66 ⅔% more ‘Laddu’ on Sunday than that of Saturday. Total number of Barfi prepared on Sunday and Saturday are equal to total number of Laddu prepared in these two days. Barfi prepared on both of days are equal in quantity then find number of Laddu that remained unsold on Sunday, given that difference between total sold Barfi to total sold Laddu in these two days is 100kg. (a) 2400 (b) 1200 (c) 900 (d) 1600 (e) None of these Q107. Ratio of Laddu prepared on Friday and Laddu sold on Monday together to total number of Laddu prepared on Friday and Monday is 113 : 120. He earns profit of Rs. 20/kg on selling Laddu and no loss on unsold Laddu. If total profit earned on Monday is Rs. 11040 more than that of Friday on selling Laddu then find quantity of Laddu prepared by him on Monday. (a) 1680 kg (b) 1600 kg (c) 1800 kg (d) 1512 kg (e) 1200 kg Q108. Quantity of Laddu sold on Friday is equal to quantity of Barfi sold on Monday. Calculate the quantity of Barfis prepared on Friday, if he prepared 80 kg more Laddu than Barfi on each day(Friday and Monday) and Barfi’s prepared on Friday is 30% less than Barfi’s prepared on Monday. (a) 1000 kg (b) 1600 kg (c) 1120 kg (d) 1200 kg (e) 1680 kg Q109. If he earns a profit of Rs. 10/kg on selling each sweets and loss of Rs. 10/800 gm on unsold items. Find his approximate profit % on Saturday, if it cost Rs. 200/kg to prepare each sweets and ratio of Laddu prepared to Barfi prepared is 5 : 4 on Saturday. (a) 2% (b) 5% (c) 10% (d) 12% (e) can’t be determined Directions (111-112): The daily work of two men is equal to that of 3 women or that of 4 youngsters. By employing 14 men, 12 women, and 12 youngsters a certain work can be finished in 24 days. Q111. If it is required to finish it in 14 days and as an additional labor, only men are available, how many of them will be required? (a) 20 (b) 30 (c) 25 (d) 15 (e) None of these 1 Q112. If it is required to finish it in 195 days and as an additional labor, only women and youngsters are available in pairs, how many pairs of women and youngsters will be required? (a) 7 (b) 5 (c) 6 (d) 8 (e) None of these Directions (113-115): A article is mark up above cost price such that markup percent is double of the profit percent. If 1 discount is 12.5%, then profit percent increased by 33 3 %. Q113. On selling 20 such article, profit is Rs.300. Find the M.P. of each article. (a) Rs.60 (b) Rs.160 (c) Rs.80 (d) Rs.240 (e) Rs.72 Q114. If shopkeeper cheat his customer by giving 20% less quantity and reducing value of discount percentage by 20% then find the new profit percent. (a) 60% (b) 75% (c) 62.5% (d) 80% (e) 70% Q110. Find the approximate average % of sweets sold on Saturday, Sunday and Monday together if shopkeeper prepared same quantity of Barfi on these days and ratio of Barfi to laddu prepared in these 3 days is 4 : 3, 4 : 5 and 20 : 21 respectively. (a) 85 (b) 76 (c) 60 (d) 68 (e) 65 16 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q115. Initially shopkeeper have 20 articles. Out of 20, 7 articles damaged and remains unsold. Marked Price should he labeled by how much percent more than cost price so that his overall profit does not change neither his discount percentage. (a) 156% (b) 146% (c) 136% (d) 120% (e) 125% Direction (116–120): Data given below shows farmers producing three types of grains. Study the data carefully and answer the following questions. Farmers producing CORN is 20% more than farmers producing RICE while number of farmers producing only CORN and only RICE is same. Ratio between farmers producing only WHEAT to farmers producing rice is 3 : 5. Farmers producing WHEAT and CORN but not RICE is twice of farmers producing only WHEAT and RICE. Number of farmers producing only RICE is half of number of farmers producing RICE. Number of farmers producing RICE and CORN is 12% of total number of farmers while number of farmers producing CORN is 48% of total number of farmers. Number of farmers producing all three grains is 16, which is 8 less than number of farmers producing only RICE and WHEAT but not CORN. Q116. Find number of farmers producing at least two types of grain are what percent of number of farmers producing at most one type of grain? (a) 37.5% (b) 43.75% (c) 50% (d) 56.25% (e) 62.5% Q117. Number of farmers producing Sugarcane is 25% more than number of farmers producing WHEAT and CORN. Find number of farmers producing Sugarcane? (a) 80 (b) 50 (c) 60 (d) 45 (e) None of the given options Q118. Find what percent of farmers producing WHEAT out of total farmers? (a) 24% 1 (b) 56 % 3 1 (c) 53 % Q120. Find the ratio between number of farmers producing only two types of grains to farmers producing all three grains? (a) 27 : 4 (b) None of the given options (c) 23 : 16 (d) 23 : 8 (e) 23 : 4 Directions (121-125): Pie-chart shown below shows percentages of markers sold by six sellers. Table shows ratio three type of marker out of total markers sold by different sellers. Study the data carefully and solve the following questions. Total Markers sold = 15,000 A 12% F 20% B 15% E 21% C 18% D 14% Type of markers A B C D E F X 4 3 9 6 3 4 Y 3 4 7 4 2 5 Z 2 3 9 5 1 3 Q121. Seller ‘A’ fixed his selling price of markers at 40% above the cost price but at the time of selling he gave 40%, 20% and 10% discount on X, Y and Z respectively. Find the total profit or loss percentage if cost price of all the markers is same? 1 (a) 2 3 % 2 (b) 1 3 % 1 (c) 3 % 3 2 3 (d) 2 3 % (d) 40% (e) 48% 17 Q119. Find total number of farmers producing either only RICE or only WHEAT? (a) 120 (b) 132 (c) 280 (d) None of the given options (e) 138 1 (e) 1 3 % www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q122. Seller ‘E’ and ‘F’ keep the S.P. of each X, Y and Z markers same and total S.P. of X, Y, Z sold by E is Rs.47250. Find the total S.P. of all the markers sold by F if E kept the SP of each X, Y, Z marker in the ratio 1 : 1.5 : 3. (a) Rs.48250 (b) Rs.51250 (c) Rs.54520 (d) Rs.57520 (e) Rs.45500 Q126. A, C and D start working together but due to bad health 1 1 of A and D their efficiency decreased by 12 2 % and 33 3 % respectively. Then find in how many hours total task will be completed by these three? 1 (a) 22 hours 2 1 (b) 10 4 hours 1 (c) 12 4 hours 1 Q123. Seller ‘C’ sold all the markers for a certain sum and 1 there was a loss of 11 9 %. Had it been sold for Rs.9000 more, (d) 9 hours 4 1 (e) 13 hours 4 1 there would have been a gain of 11 9 %. If seller ‘C’ wants to earn 20% profit, then what would be the total S.P. of Y marker if S.P. of each marker is in the ratio 2 : 3 : 4 respectively. (a) Rs.13680 (b) Rs.12680 (c) Rs.13608 (d) Rs.12608 (e) None of these Q124. There are two customers, Satish and Veer. Seller ‘B’ sells 60% of X marker to Satish, and remaining to Veer, B also sells 40% of Y marker to Satish and remaining to veer. Find the S.P. of each Y marker if Satish and Veer pays Rs.8370 and Rs.9180 for X and Y marker respectively. (a) Rs.10 (b) Rs.12 (c) Rs.14 (d) Rs.16 (e) Rs.18 Q125. Out of six sellers, which seller sells maximum number of X type of marker? (a) B (b) C (c) D (d) F (e) None of these Direction (126 - 130): Given below bar graph show number of hours taken by six persons to complete a task individually. Read the data carefully and answer the questions: 100 90 Q127. E and F start working together on another task, while F work with 25% less efficiency. E and F work for y hours and remaining work complete by B in (y + 1) hours, if ratio of work done by E and F together and by B alone is 2 : 1, then in how many hours A will complete same task alone? 1 (a) 15 hours 2 1 (b) 13 2 hours 1 (c) 17 2 hours 1 (d) 11 2 hours 1 (e) 92 hours Q128. If G can do 50 % more work in one hour as A can do in one hour, while H can do 25% less work in one hour as B can do in one hour. C start working alone and after some time he left the task, if remaining task complete by G & H together in 23.5 hours more than C work alone. Then find total time in which work completed? (a) 32.5 hours (b) 30.5 hours (c) 28.5 hours (d) 22.5 hours (e) 16.5 hours Q129. A, B, E and F work together in first hour, while C & D together destroy the task (with same efficiency of completing the task) in second hour. If this rotation continue till the total work is completed. Find how many hours required to complete the task? 17 (a) 55 hours (b) 45 80 25 17 hours 25 177 70 (c) 40 245 hours 60 (d) 50 50 (e) 59 802 hours 40 30 20 10 0 A 18 B C D E www.bankersadda.com F | 705 802 705 hours Q130. If E work for 12 hours, B work for 35 hours, then find in how many hours remaining work will be completed by C? (a) 8 hours (b) 10 hours (c) 12 hours (d) 15 hours (e) 20 hours www.sscadda.com | www.careerpower.in | Adda247 App Directions (131-135): Table given below shows length of six train, speed of train (meters/minutes), time taken by different trains to cross different platform and length of each platform is also given. Some of the data in table is missing, calculate the missing data and answer the questions according to condition given in questions. Train A B C D E F Length of train (m) ––– 180 ––– 120 300 ––– Speed (meters/ minutes) 750 ––– 2000/3 ––– ––– 1000 Time taken by train to cross platform (sec) 24 21.6 ––– ––– 30 ––– Length of platform (m) ––– ––– ––– 240 ––– ––– Q135. Two train B and D moving in same direction, if speed of smaller train is 54 km/hr and faster train crossed a man, who sits in smaller train in 24 sec. Then find the speed of faster train in meters/sec? (a) 35/2 sec (b) 45/2 sec (c) 33/2 sec (d) 43/2 sec (e) 41/2 sec Directions (136-140): Table given below shows number of lectures taken on different days in four different weeks of a month. Study the data carefully and answer the following questions. Q131. Train A running at its average speed crossed the 9 platform and it takes 911 sec to pass a man who is walking at 10 km/hr in the opposite direction to that of train A. Find the length of platform? (a) 300 m (b) 150 m (c) 180 m (d) 200 m (e) 225 m Q132. Ratio of length of train C to train F is 3 : 5. If running in opposite direction, both train crossed each other in 14.4 sec. then find time taken by train C in crossing a platform which is 50m more than length of train of C. Also find length of train F? (a) 53/2 sec, 250m (b) 60/7 sec,200m (c) 63/2 sec,250m (d) 43/2 sec,180m (e) 49/2 sec, 225 m Q133. If train B and E crossed their respective platforms and platform lengths are same as their respective train. Then find in what time faster train crossed slower train if they are running in same direction? (a) 144 sec (b) 134 sec (c) 140 sec (d) 240 sec (e) 225 sec Q134. Ratio of magnitude of time taken by train D in crossing the respective platform to speed of train D (m/s) is 5 : 8. Then find ratio of time taken by train D to train F in crossing a platform whose length is 600 m? (Ratio of length of train D to that of train F is 3:5) (a) 5 : 8 (b) 5 : 6 (c) 7 : 5 (d) 2 : 5 (e) 5 : 3 19 www.bankersadda.com | Lectures on Monday and Tuesday each Lectures on Wednesday and Thursday each Lecture on Friday and Saturday each First week Second week Third week Fourth week Amount paid for one lecture per hour 8 — 6 9 Rs. 4,000 7 5 — 8 Rs. 6,000 6 6 8 10 Rs. 9,000 (i) There are no lectures on Sunday. (ii) Month = First week + Second week + Third week + Fourth week Q136. A professor gets Rs. 2 lakhs to take all lectures on Monday and Tuesday together in that month. If lectures are of 45-minute, 30-minute, 1 hour 40 minute and 40 minutes in first, second, third and fourth week respectively then find number of lectures taken on Monday of second week? (a) 4 (b) 6 (c) 3 (d) 12 (e) 8 Q137. Find amount obtained by a professor who takes lectures in whole fourth week having 40 minutes class on each day of that month. (a) 3.48 lakh (b) 1.74 lakh (c) 2.32 lakh (d) 1.16 lakh (e) None of these www.sscadda.com | www.careerpower.in | Adda247 App Q138. A professor takes lectures only on Wednesday. Find lectures taken by him on third week is what percent more or less than lectures taken by him on second week if he earns Rs. 1,21,500 by taking 45 minutes lecture on each week of that month. (a) 100% (b) 60% (c) 80% (d) 20% (e) 40% Q139. Find the ratio between amount given to professor who takes 45 minutes lecture on Wednesday of second week to professor who takes 40 minutes lecture on Saturday of fourth week. (a) 1 : 2 (b) 3 : 8 (c) 1 : 3 (d) 3 : 5 (e) 8 : 27 Q140. Amount given to a professor who takes lecture on whole second week [1 hour each] is 12.5% more than amount given to that professor who takes lecture on whole first week [45 minutes each]. Find numbers of lectures taken on Monday of second week? (a) 8 (b) 4 (c) 6 (d) 12 (e) None of these Directions (141–145): Line chart given below shows markup % above C.P. and discount % given on article X on five different days while table given below shows ratio between mark price of article X to mark price of article Y on these days. Study the date carefully and answer the following questions. Mark up % Discount % 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% Days Monday Tuesday Wednesday Thursday Friday 20 Q141. Ratio between cost price of article X on Monday to Thursday is 4 : 5. Retailor earns Rs 75 profit on selling article X on Thursday. Find profit earned on selling article Y on Monday, if he give 35% discount on marked price of Y while Cost price of article X and Y is same on Monday. (a) Rs 465 (b) Rs 444 (c) Rs 525 (d) Rs 555 (e) Rs. 585 Q142. On selling article X and Y on Friday, retailor earns total Rs 1430 profit. If selling price of article Y is Rs. 520 more than selling price of article X and cost price of article X and Y is same, then find discount percent offered on article Y. 1 (a) 33 3 % 2 (b) 16 % 3 2 (c) 66 % 3 (d) 25% (e) 12.5% Q143. Ratio of cost price of article X to article Y on Wednesday is 3 : 4. Retailor earn 26% profit on selling article Y by giving discount of Rs. 1008. Find the difference between selling price of both articles. (a) Rs. 900 (b) Rs. 540 (c) Rs. 360 (d) Rs. 1080 (e) Rs. 1260 Q144. Find profit percentage earned by retailor on selling article Y on Tuesday at 40% discount if cost price of article Y is 20% less than that of cost price of article X? (a) 60% (b) 25% (c) 50% (d) 75% (e) 100% Mark price of X : Mark price of Y 4:5 3:4 11 : 14 7:9 5:7 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q145. If profit earned on selling article X on Monday is 176 then find the total profit earned by retailor on selling article X on all five days. It is given that ratio between cost price of article X on respective days starting from Monday to Friday is 1 : 2 : 3 : 4 : 5. (a) None of the given as option (b) Rs. 1432 (c) Rs. 1260 (d) Rs. 1302 (e) Rs. 1424 Directions (146-150): The following information is about the production of bikes by 3 different companies from Monday to Friday in a specific week. Read the information carefully and answer the following question: The total production by 3 companies on Monday was 540 out 1 of which 33 3 % bikes were produced by Hero. The number of bikes produced by Bajaj on Monday are less than the bikes produced by Hero on Monday by the same extent as the number of bikes produced by Honda on Monday is more than the bikes produced by Hero on Monday. The difference between bikes produced by Bajaj and Honda on Monday is 40. 150 bikes are produced by Hero on Tuesday, which is 100 less than the bikes produced by the same company on Wednesday. A total of 910 bikes were produced by Hero from Monday to Friday. The ratio between bikes produced by Hero on Thursday to bikes produced by the same company on Friday is 5 : 6. 220 bikes were produced by Bajaj on Tuesday, which is 80 less than the bikes produced by Honda on Wednesday. A total of 570 bikes were produced on Tuesday, which is 76% of the total bikes produced on Wednesday. The number of bikes 2 produced by Honda on Thursday is 66 3 % more than bikes produced by Hero on the same day. Total 580 bikes were produced on Thursday. The number of bikes produced by Honda on Friday is same as that on Monday. 140 bikes were produced by Bajaj on Friday. Q146. Find the ratio between total bikes produced on Monday to that on Wednesday. (a) 18 : 29 (b) 18 : 25 (c) 18 : 31 (d) 3 : 5 (e) None of these Q147. Find the total number of bikes produced by Bajaj from Monday to Friday. (a) 900 (b) 980 (c) 950 (d) 960 (e) None of these 21 www.bankersadda.com | Q148. Find the average number of bikes produced per day by Honda from Monday to Friday. (approximate) (a) 250 (b) 220 (c) 270 (d) 240 (e) 230 Q149. On which pair of days out of the following, the number of bikes produced by Hero is the same ? (a) Tuesday and Wednesday (b) Wednesday and Thursday (c) Tuesday and Thursday (d) Monday and Wednesday (e) Monday and Tuesday Q150. On which day the total number of bikes produced was the maximum ? (a) Monday (b) Tuesday (c) Wednesday (d) Thursday (e) Friday Q151. Ram spends Rs. 8000 for 4 years in scheme ‘A’ which offers R% p.a. at S.I. and Rs. 10,000 for 2 years in scheme ‘B’ which offers R% p.a. at C.I. He got same amount from both schemes. Find value of ‘R’. I. 20% II. 50% III. 100% (a) Either I or II (b) Either I or III (c) Either II or III (d) Any one of I, II and III (e) Only I Q152. A retailer bought some eggs at Rs. 10 each. Out of these, some are rotten, and some are fresh. He sold fresh eggs at 50% profit and rotten eggs at 20% loss. In total he earns 8% profit. Find the probability of selecting one fresh egg from total eggs. 1 (a) 9 1 (b) 2 (c) (d) 4 5 3 5 2 (e) 5 Q153. Two equilateral triangles and two parallelograms are cut down from a hexagon. If area of one equilateral triangle is 4√3 cm² then find the height of parallelogram? (a) 2√3 cm (b) √3 cm (c) 4√3 cm (d) 3√3 cm (e) 8√3 cm www.sscadda.com | www.careerpower.in | Adda247 App Q154. Work done by A, B, C and D is in arithmetic progression. C worked for 9 days and completed 30% work. A and B worked for 2 days and 3 days only. Find for how many days ‘D’ work to complete the remaining work if all four together can complete that work in 4 days. (a) 12 days (b) 10 days (c) 8 days (d) 6 days (e) 4 days (a) 12100 (b) 14400 (c) 13800 (d) 15000 (e) None of these Directions (155-156): Pipe A is 33 ⅓% more efficient than pipe B. Pipe C can empty the tank at the rate of 40 litres/hour. 18 11 % less of their respective efficiency respectively and B Q155. Mohit opened pipe A and pipe B for 13 7 hours and the tank was filled upto10 liters less than 75% of its capacity. After that he also opened the pipe C and tank was filled in 1 ℎ𝑜𝑢𝑟. Find the time in which pipe C alone could empty the tank. (a) 8 hours (b) 9 hours (c) 9.5 hours (d) 8.5 hours (e) Can’t be determined Q156. When these pipes were connected to a different tank. Initially, pipe A and B were opened, till the tank was filled 10 liters more water than 75% capacity of tank. After that all the pipes were opened, and the remaining tank was filled in 1 hour. Find the capacity of tank. Given that pipe A will take 4 hours to fill the tank. (a) 600 litres (b) 160 litres (c) 360 litres (d) 180 litres (e) can’t be determined Q157. Speed of current is 10 km/hr and speed of a motor boat is 80% more than speed of current. Motor boat travels 280 km downstream with its usual speed, after that it’s increased speed by ‘s’ kmph and travelled for another 280 km then it returns and covers 560 km in upstream. If boat complete whole journey downstream to upstream in 45 hr, then find the value of ‘s’? (a) 10 km/hr (b) 8 km/hr (c) 6 km/hr (d) 12 km/hr (e) 4 km/hr Q158. A, B and C enter into a partnership and invested some amount. After one year A double its investment, B increase its 1 investment by 33 3 % and C increase its investment by 20%. In the third year A and B withdraw their investments and D joins the partnership with C. After three year they got profit in the ratio of 12 : 14 : 17 : 8 (A : B : C : D). If difference between initial investment of B and C is 1150. Then Find out the total initial investment made by A and D together? 22 www.bankersadda.com | Q159. A and B together can finish first work in 1 11 4 𝑑𝑎𝑦𝑠 while A alone can do it in 25 days. Both work on 2 second work for ‘x’ days and (x + 4) days with 22 9 % more and 2 gets Rs 180 more wages than that of A out of total wages Rs 4140. Find the time taken by both together to complete the second work if they work with their original efficiency? 4 (a) 13 5 𝑑𝑎𝑦𝑠 1 (b) 12 𝑑𝑎𝑦𝑠 2 2 (c) 16 𝑑𝑎𝑦𝑠 3 1 (d) 13 𝑑𝑎𝑦𝑠 3 (e) 14 𝑑𝑎𝑦𝑠 Q160. A shopkeeper raised the marked price of an article by 125% and allows three successive discounts of 25%, 20% and 1 119 % on new marked price and make a profit of 25%. If 1 shopkeeper allows three successive discounts of 33 3 %, 2 1 16 3 % and 122 % on new marked price, he would have Rs 255 less profit than earlier, then find cost price of that article? (a) 2184 Rs. (b) 2244 Rs. (c) 2284 Rs. (d) 2304 Rs. (e) 2364 Rs. Directions (161-165): The following table shows different plans offered by a lender, type of interest and rates of interest applicable during first, second and third years. (Note: Some values are missing, you need to calculate those values if required.) Rate of Interest Type of Plans First Second Third Interest Year Year Year A Simple 2 2 6 3% 3 3% Interest B Compound 1 6 % 4 Interest C Simple 3 1 8 % 5 % 4 4 Interest D Compound 1 3 7 2% 4 4% Interest 4 3 E Simple 5 % 4 % 5 5 Interest www.sscadda.com | www.careerpower.in | Adda247 App Q161. If two persons borrows an equal amount of Rs.12000 under plan B and plan E respectively and rate of interest for the first year under plan B and D is same, then what is the difference between second year’s interests alone paid by each of them? (a) Rs.105.25 (b) Rs.110.25 (c) Rs.115.25 (d) Rs.120.25 (e) Cannot be determined Directions (166-170) : A new party XYZ, participated in election in 5 constituencies (A, B, C, D & E). Pie chart shows % distribution of party’s total votes scored by its 5 candidates and line graph shows, % of votes scored by winner candidates out of total votes polled in these 5 constituencies. Note: All the votes must be calculated in nearest hundreds. TOTAL VOTES = 300000 A 10% E 12.5% Q162. A person borrows Rs.20480 under plan C. After completion of the loan tenure of three years under plan C, he extends the tenure for further two years under plan D on the amount payable at that time. He settles his loan by paying Rs.27778. What is the rate of interest for the second year under plan D if rate of interest for the third year under plan C and D is same? 3 (a) 5 % B 25% D 15% 4 1 (b) 5 4% C 37.5% 1 (c) 6 4% 3 (d) 4 4% 3 (e) 6 4% 70 Q163. If the amounts borrowed by a person under plan B and C are in the ratio 16 : 13 and rate of interest applicable during the first year under plan B and D is same, then what is ratio of interests payable under these plans at the end of second year. (a) 5 : 6 (b) 3 : 5 (c) 3 : 4 (d) 5 : 4 (e) None of these Q164. The lender decides to offer a fixed rate of interest at 2 6 3% per year under plan C. By how much percent the interest payable will increase from the interest payable previously under the old plan for the period of three years if rate of interest for the third year under old plan C and plan D is same? 1 (a) 6 3% 1 (b) 6 4% 2 (c) 6 % 3 2 (d) 6 5% (e) Cannot be determined Q165. Rates of interest for the first year under plan A and E 2 3 are 8 3% and 7 5% respectively. A person borrows a total of Rs.30000 partially under plan A and E and pays a total interest of Rs.5540 at the end of third year. How much amount does he borrow under plan A? (a) Rs.14000 (b) Rs.18000 (c) Rs.16000 (d) Rs.12000 (e) Rs.20000 23 www.bankersadda.com | 60 50 40 30 20 10 0 A B C D E Q166. Candidate of party XYZ is winner from constituency B and defeated runner-up candidate by 7000 votes. None of the candidates from this constituency obtained vote less than 12000, then calculate maximum possible number of candidates from constituency B. (a) 6 (b) 7 (c) 15 (d) 13 (e) can’t be determined Q167. Party XYZ wins from constituency C. In constituency E total votes polled are 45000 less than that of constituency C. Find from how many votes the candidate of party XYZ lost in constituency E? (a) 79500 (b) 91500 (c) 26500 (d) 27000 (e) 0 www.sscadda.com | www.careerpower.in | Adda247 App Q168. Out of constituency A and D, one of the seats was won by party XYZ. In lost constituency their candidate was runner up by 15000 votes. Find maximum possible difference between total votes polled in both of these constituencies. (a) 50000 (b) 37500 (c) 25000 (d) either (a) or (b) (e) either (b) or (c) Q169. In constituencies A & E, both candidates of party XYZ are runner up, and lost by same number of votes. If the votes polled in E is 180000, find approximate number of votes polled in constituency A. (a) 350000 (b) 365000 (c) 180000 (d) 100000 (e) None of these Q170. Which of the following condition is never possible? (i) constituency B and C have equal number of votes polled given that party wins from C. (ii) candidate of party XYZ lost by 28000 voters from constituency E. (iii) constituency A and B have equal number of votes polled. (iv) winner candidate of constituencies B and D got equal number of votes. (a) only (i) (b) Only (i) and (iii) (c) (i), (ii) and (iv) (d) (i) and (ii) (e) All of these are possible Directions (171-175): Data show the different kind of solids in a toy shop. Shopkeeper or (toymaker) makes different types of toys by joining these solids. Some values are missing, you have to calculate these values if required to answer the question. Diameter Length Breadth Height Cylinder – – – 12 Cube – – – – Cuboid – 24 – 10 Cone 14 – – – Sphere 21 – – – Hemisphere − – – – Q171. A toymaker makes a toy in which a cone is mounted on the base of a hemisphere. If the total surface area of the toy is 858 cm² then find the volume of the toy? 2 (a) 1950 3 cm² 2 (b) 1250 3 cm² (c) 1400 cm² (d) 1500 cm² (e) None of these 24 www.bankersadda.com | Q172. Toymaker mounted the cube on the cylinder such that cylinder top is exactly in the middle of the face of the cube. Find the total surface of the toy formed, if the height of formed 11 toy is 4 𝑡ℎ of the height of cylinder and curved surface area of cylinder is 66 times the height of cylinder. (a) 3125 cm² (b) 2794.5 cm² (c) 4112 cm² (d) 5123 cm² (e) 3438 𝑐𝑚2 Q173. If given sphere is cut into two hemisphere and these hemispheres are mounted on both ends of the cylinder, then find out the ratio of volumes of toy formed by joining both hemispheres on cylinder, cylinder and sphere. (a) 7 : 6 : 13 (b) 6 : 13 : 7 (c) 13 : 6 : 7 (d) 13 : 7 : 6 (e) None of these Q174. Volume of the cuboid is approximately what percent more or less than the volume of cone if slant height of cone is 25 cm and the breadth of the cuboid is 25% of the height of cone. (a) 7% (b) 11% (c) 14% (d) 17% (e) 21% Q175. A solid right circular cylinder has radius r and height 5r. A solid right circular cone is carved out from one end of the base of cylinder. If base radius of cone is r and height is 2√2 r then, find the ratio between total surface area of cone to the total surface area of remaining part of cylinder. (a) 3 : 5 (b) 4 : 7 (c) 2 : 7 (d) 3 : 4 (e) 1 : 3 Directions (176-180): Table given below shows the amount paid, mileage and time (in hours) for which five taxis are hired by a business man at the rate of Rs. 5/km plus the cost of petrol at Rs. 60/Litre. Taxi P Q R S T www.sscadda.com Amount paid (in Rs.) 2640 3500 2592 4500 2925 | Mileage (in km/Litre) 10 12 15 18 14 www.careerpower.in | Time (hours) 16 25 18 30 15 Adda247 App Q176. Amount paid to taxi Q and S together will how much increase, if mileages of both taxis are decreased by 2km/Litre (distance covered will be same. (a) Rs.470 (b) Rs.475 (c) Rs.575 (d) Rs.674 (e) Rs.325 Q177. Average speed of taxi Q is what percent less/more than the average speed of taxi P? 1 (a) 13 % 3 2 (b) 12 % 3 (c)10% 2 (d) 6 3 % 2 (e) 16 3 % Q178. If taxi S travel his journey in two equal halves with two different speed having ratio 3:2 then find the time for which it travels with higher speed? (a) 12 hours (b) 10 hours (c) 16 hours (d) 18 hours (e) 14 hours Q179. What is the ratio of the average speed of the taxi R to that of taxi T? (a) 21:16 (b) 12:23 (c) 14:17 (d) 16:21 (e) None of these Q180. Find the average of the total distance covered by all five taxis? (a) 314.6 km (b) 298.4 km (c) 412 km (d) 344.6 km (e) 346.6 km Q181. A mixture contains milk and water in the ratio of 9 : 2. 44 lit of mixture is taken out and 12 lit of water is added to it, such that ratio of milk to water becomes 3:1. Now another mixture of 64 lit having milk and water in ratio of 3:5 is added to it. Find ratio of milk to water in the final mixture? (a) 32 : 17 (b) 34 : 19 (c) 7 : 4 (d) 24 : 11 (e) 33 : 19 25 www.bankersadda.com | Q182. Ratio of present age of Simmi & Rimmi is 4 : 3 while ratio of age of Simmi 6 years later to present age of Rina is 3 : 1. Ratio of present age of Rimmi to present age of Rina is 2 : 1. Find average of present age of all the three (in years). (a) 37 (b) 36 (c) 32 (d) 34 (e) 38 Q183. A shopkeeper sells an article at Rs. 720 by making a profit of x% and on interchanging selling price with cost price, it would have a loss of y%. Find the selling price if an article was sold at y% profit whose cost price is Rs.720 (given x : y = 9 : 7) (a) Rs. 810 (b) Rs. 800 (c) Rs. 1100 (d) Rs. 880 (e) Rs. 660 Q184. There are two rectangular fields of same area. The length of first rectangular field is a% less than the length of the second field and breadth of the first field is (4a)% greater than the breadth of the second field. Find the value of ‘a’ if a is non-zero number. (a) 60 (b) 75 (c) 80 (d) 90 (e) None of these Q185. A sum is divided between Aman and Vikash in the ratio of 3 : 5. Aman purchased a motorbike from his money which depreciates 20% per annum while Vikash invested his amount in a scheme which offers compound interest at 20% per annum. By what percentage, the total sum will change after two years? (a) 12% (b) 14% (c) 16% (d) 18% (e) 20% Directions (186-190): Read the following table and line graph carefully and answer the following questions. Following table shows the time taken by five persons to complete a work on Monday and Ratio of Time taken by these five persons to complete the work on Monday to the time taken to complete the work on Wednesday is also given. Line graph shows the efficiency (as a percentage) of these five persons on Tuesday with respect to that on Monday. www.sscadda.com | www.careerpower.in | Adda247 App Person Time taken to complete the work on Monday Ratio of Time taken to complete the work on Monday to the time taken to complete the work on Wednesday Gaurav 25 min. 5:4 Abhishek 20 min. 4:5 Shailesh 50 min. 10 : 7 Neeraj 10 min. 5 : 13 Arunoday 150 min. 3:5 Q188. On Tuesday, Abhishek, Shailesh and Neeraj work in a rotation in this order to complete the job with only 1 person working in a minute. They earned a total of 875 Rs. Find the share of Shailesh. (a) 41 Rs. (b) 31 Rs. (c) 51 Rs. (d) 49 Rs. (e) None of these Q189. On Tuesday, Aman who is half as efficient as Shailesh, worked for 50 minutes on the same day then he left. In how many minutes Neeraj and Abhishek together will complete the remaining work ? 120 2 (a) 5 9 mins. 80 (b) 4 7 mins. percentage → 100 3 3 (c) 5 mins. 60 7 1 (d) 4 7 mins. 40 5 (e) 5 7 mins. 20 0 Person → Q186. Gaurav, Abhishek and Neeraj work in a rotation to complete the job on Tuesday with only 1 person working in a minute. Who should start the job so that the job is completed in the least possible time? (a) Gaurav (b) Abhishek (c) Neeraj (d) Any one of three (e) Can’t determine Q187. On Tuesday, Gaurav and Arunoday started the work and they worked for 5 minutes then Gaurav is replaced by Abhishek. In how many minutes Abhishek and Arunoday complete the remaining work ? 3 (a) 20 7 min. (b) 21 (c) 21 4 21 5 21 4 min. min. (d) 20 17 min. (e) None of these 26 www.bankersadda.com | Q190. On Wednesday, all of them started the work together. After working for 2 minutes Gaurav left. All except Gaurav worked for another 3 minutes and then all left except Arunoday. In how much time Arunoday will complete the remaining work? (find the approximate value) (a) 86 minutes (b) 81 minutes (c) 96 minutes (d) 56 minutes (e) 79 minutes Directions (191-195): Each of the following question is followed by two quantities I, and II. You have to determine the value of the quantities using the information provided and accordingly compare the quantities. Mark your answer as per the instruction set provided below. (a) Quantity I>Quantity II (b) Quantity I≥Quantity II (c) Quantity I<Quantity II (d) Quantity I≤Quantity II (e) Quantity I=Quantity II or no relation Q191. Quantity I: Time taken by Prabhas to cover a distance of 200 km (without stoppage) by car through which he covers a distance of 100 km in 1 hour when he stops to get fuel for 10 minutes. Quantity II: Time taken by Nagarjuna to travel 20 km downstream if speed of boat and speed of stream be 7 kmph & 3 kmph respectively. www.sscadda.com | www.careerpower.in | Adda247 App Q192. There are 3 red and 5 blue balls in the urn. Quantity I: the probability of drawing 1 red ball and 2 blue balls. Quantity II: the probability of drawing 2 red balls and 1 blue ball. Q198. A bag contains total 12 balls in which there are 5 green balls and rest are blue and red balls. What is difference between blue & red balls. Statement I . If one ball taken out from bag probability of 7 being either red or blue is 12. Statement II. If two balls taken out from bag probability of Q193. In how many ways can 4 boys or 5 girls be selected? Quantity I: there are 20 persons (boys and girls only) in the group out of which 12 are boys. Quantity I: the group comprises 10 boys and 10 girls. Q194. Quantity I: The length & breadth of a rectangle of perimeter 48 cm are in ratio 5:3. Find Area? Quantity II: 135 1 1 𝑎 𝑏 Q195. + =? Quantity I: a & b are roots of equation 𝑥 2 + 𝑥 − 6 = 0 Quantity II: 𝑎2 + 𝑏 2 = 20; 𝑎𝑏 = 8 Direction (196-200): Following are the questions based on two statements and answer the following based on the given statements. Q196. What will be respective ratio of saving of Veer & Deepak. Statement I. Income of Veer is 4% less than that of Sameer and also expenditure of Veer is 12.5% less than that of 3 Sameer. Deepak spend 5 th of his income. Statement II. Sameer save Rs. 7000 & Veer save Rs. 7400. Income of Deepak is Rs. 1000 more than that of Sameer. (a) Only statement I is sufficient (b) Only statement II is sufficient (c) Statement I and II both together is sufficient (d) Either statement I or Statement II alone is sufficient (e) Neither statement I nor statement II is sufficient 1 being either red or blue is 6. (a) Only statement II is sufficient (b) Either statement I or Statement II alone is sufficient (c) Statement I and II both together is sufficient (d) Only statement I is sufficient (e) Neither statement I nor statement II is sufficient Q199. Side of square is 3.5 cm more than radius of circle. What will be area of square? Statement I. Difference between circumference and diameter of circle is 45 cm. Statement II. Radius of circle is 50% more than breadth of rectangle whose length is 15 cm. Ratio of circumference of circle & perimeter of rectangle is 3 : 2. (a) Only statement II is sufficient (b) Either statement I or Statement II alone is sufficient (c) Statement I and II both together is sufficient (d) Only statement I is sufficient (e) Neither statement I nor statement II is sufficient Q200. What will be length of train A? Statement I. Relative speed of train A & B is 10 meters/sec when both running in same direction and length of train B is 240 (Speed of train B is more than speed of train A). Statement II. Train B cross a pole in 8 sec and cross train A in 12 sec running in opposite direction. (a) Only statement II is sufficient (b) Either statement I or Statement II alone is sufficient (c) Neither statement I nor statement II is sufficient (d) Only statement I is sufficient (e) Statement I and II both together is sufficient Q197. What will be cost price of article, which marked 40% above. Statement I. If article sold 25% discount profit will be Rs. 50. 2 Statement II. If article sold two successive discounts of 14 % 7 and 10% profit will be Rs. 80. (a) Only statement I is sufficient (b) Either statement I or Statement II alone is sufficient (c) Statement I and II both together is sufficient (d) Only statement II is sufficient (e) Neither statement I nor statement II is sufficient 27 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Directions (201-215): What will come in place of question mark (?) in the following questions. Q201. √5776 - √1444 + √729 =43 + ? (a) 25 (b) 20 (c) 26 (d) 24 (e) 22 Q208. 73823–34156+4756+6758–9849 = 41499–160? (a) 5 (b) 7 (c) 4 (d) 8 (e) 6 5599 3773 (a) 44 (b) 46 (c) 48 (d) 50 (e) 52 Q202. 78 ×26÷6 +1262= 1311 + (?)2 (a) 17 (b) 22 (c) 15 (d) 13 (e) 19 1 7 (a) 2 (b) 1 (c) 0 (d) 3 (e) 4 3840 1440 1330 Q211. √ 60 + 40 − 70 = ? Q204. 42.5×15 +37.5× 25= 1420 + ? (a) 145 (b) 165 (c) 155 (d) 170 (e) 185 (a) 10 (b) 9 (c) 8 (d) 7 (e) 11 Q205. 2450 +3760 -3830 =6000 - ? (a) 3610 (b) 3620 (c) 3580 (d) 3600 (e) 3520 Q212. 25×18+ 𝑜𝑓25 20 Q210. 84 × 4 ÷ 212 +? = 147 × 21 − 21 Q203.1484÷28 + 1462÷34 -12×7= ? (a) 12 (b) 14 (c) 18 (d) 16 (e) 20 4 88 Q209. 1331 × 2036 × 49 =? −62 4200 40 − 525 105 = 740 − ? (a) 200 (b) 220 (c) 190 (d) 170 (e) 150 3 1 (a) 50 (b) 45 (c) 35 (d) 30 (e) 40 Q213. 3845+4380+2640 − 5965 = (?)2 (a) 75 (b) 60 (c) 80 (d) 70 (e) 72 Q207. 55% 𝑜𝑓 900 + 70% 𝑜𝑓 1050 = ? % 𝑜𝑓 3000 (a) 41 (b) 42 (c) 43 (d) 44 (e) 45 Q214. 400 ÷ 20 × 35 + 6666 ÷ 33+ ? = 1100 (a) 180 (b) 198 (c) 195 (d) 205 (e) 200 Q206. (5 64 ) ÷ (432 − 202 + 7 𝑜𝑓 21) × (82) =? 𝑜𝑓 64 28 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q215. 28× 14.5+1680÷15+445=1000 -? (a) 27 (b) 37 (c) 47 (d) 50 (e) 40 Q222. (a) 96 (b) 126 (c) 156 (d) 196 (e) 84 Directions (216-230): what approximate value will come in place of question (?) mark: Q216. 129.89% of 1199.82 + 1249.78 ÷ 49.98 × 30.012 = ? (a) 2210 (b) 2380 (c) 2310 (d) 2530 (e) 2460 Q217. 155.9 ÷ √168.81+ (2.98)² × 39.89 = ? % of 599.92 (a) 62 (b) 78 (c) 84 (d) 52 (e) 68 Q218. √80.98 × 36.01 + 679.81 ÷ 17.01= ? + (511.98)1/3 (a) 86 (b) 78 (c) 94 (d) 52 (e) 66 Q219. 1599.85% of 139.89 + ? % of 1599.83 = 72.01 × 39.81 (a) 20 (b) 32 (c) 60 (d) 50 (e) 40 Q220. (17.012)² + (21.89)² + (8.01)² + ? = 1749.821 – 820.01 + 2210.01 (a) 2208 (b) 2256 (c) 2601 (d) 2303 (e) 2373 Q221. 307.89 + 671.93 – 39.87% of ? + 79.89% of 354.93 = (27.87)² (a) 1200 (b) 1175 (c) 1225 (d) 1250 (e) 1280 29 www.bankersadda.com 177.8 + ? | 7.98 +24.89×41.87 –15.98 % of 400 = (31.89)2 Q223. √1295.96 + √2024.93 + √1520.97 – √? = 12.93 % of 899.98 (a) 5 (b) 7 (c) 13 (d) 16 (e) 9 Q224. 349.89 + (a) 196 (b) 441 (c) 400 (d) 529 (e) 625 55.98 ×239.89 13.86 + √? = (10.98)3 Q225. 31.96 × 34.89 + 39.98 % of 3304.98 (a) 800 (b) 700 (c) 900 (d) 1000 (e) 950 √960.89 + 18.98 % of ? = Q226. 1782.011 ÷ 53.99 + 455.889 – 2346.011 × 1.011 = ? × 2.93 (a) –629 (b) –619 (c) 629 (d) 619 (e) –609 Q227. (574.99 + 7511.11 – 2768.91) ÷ (76.1 × 0.98 + 674.976 – 342.001) = √? (a) 529 (b) 49 (c) 169 (d) 289 (e) 729 1 Q228. [(√3843.9 × 9.09) ÷ (26.99)3 ] × 23.012 =?2 + 336.97 (a) 33 (b) 23 (c) 27 (d) 37 (e) 43 www.sscadda.com | www.careerpower.in | Adda247 App Q229. √(95.99) × 12.01 ÷ 17.9 + 25.899– 9.011 =(64.9– ?)% of 35.88 (a) 50 (b) 35 (c) 30 (d) 40 (e) 20 Q236. 1229.99 + 2120.09 − 3049.987 =? (a) 300 (b) 100 (c) 200 (d) 500 (e) 400 Q237. √√(99.99 + 104.99 × 5 =?÷ 8.989 Q230. 11.9 × √224.89 + 1212.09 – (1053.11 ÷ 8.9) = ? (a) 1,275 (b) 1,225 (c) 1,175 (d) 1,255 (e) 1,245 Directions (231-245): What approximate value will come in place of question mark (?) in the following questions. (You are not expected to find the exact value) Q231. 42.022% 𝑜𝑓 350.09 − 28.04% 𝑜𝑓 399.999 =? (a) 40 (b) 35 (c) 45 (d) 50 (e) 30 Q232. √(123.09 + 465.05) ÷ 11.99+? = 240.02 ÷ 1.989 (a) 93 (b) 143 (c) 133 (d) 113 (e) 123 Q233. (15.99)2 − 14.04 × 8.99+? = 154.999 (a) 30 (b) 45 (c) 35 (d) 20 (e) 25 (a) 55 (b) 15 (c) 25 (d) 35 (e) 45 Q238. 35.99 × 4.98 − 1199.99 ÷ 7.99 =? (a) 20 (b) 50 (c) 40 (d) 30 (e) 10 Q239. ?2 + 60% 𝑜𝑓 239.99 = 55% 𝑜𝑓 320.02 + 3.98 (a) 8 (b) 6 (c) 4 (d) 16 (e) 14 Q240. 524.90 + 125.05 =? × 9.99 (a) 85 (b) 75 (c) 65 (d) 55 (e) 45 Q241. √144.04 × 15% 𝑜𝑓 120.09 =? −54.99 × 3.03 (a) 401 (b) 431 (c) 341 (d) 471 (e) 381 Q234. 62.02% 𝑜𝑓 249.99 − 19.99% 𝑜𝑓 105.05−? = 110 (a) 24 (b) 16 (c) 28 (d) 34 (e) 20 Q242. 13.03 × 7+? = 30.03% 𝑜𝑓 349.99 (a) 14 (b) 18 (c) 8 (d) 20 (e) 6 Q235. 44.98% 𝑜𝑓 220.09 + 30.03% 𝑜𝑓 160.06 =?2 + 2.99 (a) 32 (b) 28 (c) 12 (d) 22 (e) 18 Q243. 32.01% of 600.02 – 19.99% of 400.04+?=859.99÷2 (a) 358 (b) 258 (c) 288 (d) 318 (e) 338 30 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App 141 Q251. 810, 820, 832, 868, 1012, 1732, 6052 (a) 6052 (b) 810 (c) 868 (d) 832 (e) 1732 279.89 Q244. 20.09 + 39.99 − √? = 10.01 (a) 36 (b) 16 (c) 4 (d) 64 (e) 100 Q245. 8.98 × 60.02 − 19.992 + 10.01% 𝑜𝑓 130.09 =? (a) 123 (b) 93 (c) 153 (d) 173 (e) 113 Directions (246-260): In each of these questions a number series is given. In each series only one number, if any, is wrong. Find out the wrong number. Q246. 28, 14, 14, 22, 42, 105, 315 (a) 28 (b) 42 (c) 315 (d) 22 (e) 105 Q252. 1024, 350, 832, 508, 704, 604, 640 (a) 1024 (b) 640 (c) 704 (d) 350 (e) 508 Q253. 190, 210, 266, 358, 486, 646, 850 (a) 646 (b) 850 (c) 486 (d) 190 (e) 210 Q254. 15, 50, 160, 370, 709, 1208, 1904 (a) 15 (b) 50 (c) 370 (d) 1208 (e) 15 Q247. 5, 7, 13, 25, 47, 75, 117 (a) 5 (b) 7 (c) 75 (d) 117 (e) 47 Q248. 288000, 24000, 3600, 300, 50, 12.5, 6.25 (a) 24000 (b) 50 (c) 12.5 (d) 3600 (e) 6.25 Q255. 120, 170, 251, 367, 522, 720, 990 (a) 120 (b) 990 (c) 522 (d) 367 (e) 251 Q256. 55, 120, 210, 338, 517, 760, 1090 (a) 120 (b) 1090 (c) 760 (d) 55 (e) 338 Q249. 120, 125, 136, 149, 166, 185, 208 (a) 120 (b) 166 (c) 149 (d) 185 (e) 208 Q250. 205, 214, 186, 250, 125, 341, −2 (a) 205 (b) 214 (c) 250 (d) 125 (e) -2 31 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q257. 110, 140, 240, 261, 365, 380, 492 (a) 240 (b) 380 (c) 492 (d) 140 (e) 110 Q267. I. 𝑥 2 − 2𝑥 − 15 = 0 II. 𝑦 2 − 15𝑦 + 56 = 0 Q268. I. 10𝑥 2 + 19𝑥 + 7 = 0 II. 5𝑦 2 + 16𝑦 + 12 = 0 Q269. I. 𝑥 2 − 20𝑥 + 75 = 0 II. 𝑦 2 + 19𝑦 + 84 = 0 Q258. 105, 106, 123, 154, 197, 255, 327 (a) 197 (b) 105 (c) 154 (d) 255 (e) 123 Q270. I. 𝑥 2 − 9𝑥 − 22 = 0 II. 𝑦 2 − 17𝑦 + 66 = 0 Q271. I. 4𝑥 2 + 19𝑥 + 15 = 0 II. 8𝑦 2 + 10𝑦 + 3 = 0 Q259. 1, 329, 638, 911, 1130, 1277, 1334 (a) 1 (b) 1334 (c) 911 (d) 1277 (e) 638 Q272. I. 𝑥 2 − 18𝑥 + 56 = 0 II. 𝑦 2 + 4𝑦 − 32 = 0 Q273. I. 𝑥 2 + 14𝑥 − 72 = 0 II. 𝑦 2 − 13 + 36 = 0 Q260. 2100, 2136, 1990, 2316, 1740, 2640, 1344 (a) 2100 (b) 1990 (c) 2316 (d) 1740 (e) 2640 Q274. I. 𝑥 2 − 92 = 122 II. 𝑦 3 = 3375 5 Q275. I. 𝑥2 28 3 = 𝑥2 7 II. 11𝑦 + (7 × 6) = 97 Direction (261-275): Given below in each question two quadratic equations are given. Please solve each quantity and compare both of them and answer accordingly from the following options. (a) x>y (b) y> x (c) x ≥y (d) x ≤y (e) x =y or No relation can’t be established. Q261. I. 2𝑥 2 + 𝑥 − 6 = 0 II. 𝑦 2 + 6𝑦 + 9 = 0 Q276. I. 𝑥 2 − 22𝑥 + 72 = 0 II. 𝑦 2 + 11𝑦 + 30 = 0 Q262. I. 𝑥 2 − 4𝑥 + 4 = 0 II. 𝑦 2 − 10𝑦 + 16 = 0 Q277. I. 𝑥 2 − 23𝑥 + 120 = 0 II. 𝑦 2 − 17𝑦 + 70 = 0 Q263. I. 2𝑥 2 + 7𝑥 + 6 = 0 II. 3𝑦 2 + 11𝑦 + 10 = 0 Q264. I. 𝑥 2 − 2𝑥 − 24 = 0 II. 𝑦 2 − 12𝑦 + 36 = 0 Q278. I. 𝑥 2 − 15𝑥 + 54 = 0 II. 𝑦 2 + 10𝑦 − 96 = 0 Q265. I. 4𝑥 2 + 11𝑥 + 6 = 0 II. 𝑦 2 + 10𝑦 + 25 = 0 Q279. I. x3+440=2168 II.y2−23=121 Q266. I. 4𝑥 2 − 20𝑥 + 25 = 0 II. 5𝑦 2 − 6𝑦 − 8 = 0 32 www.bankersadda.com Directions (276-290): In each of the following questions, two equations (I) and (II) are given. Solve the equations and mark the correct option: (a) if x>y (b) if x≥y (c) if x<y (d) if x ≤y (e) if x = y or no relation can be established between x and y. | www.sscadda.com | www.careerpower.in | Adda247 App Directions (291-295): Pie chart given below shows distribution of passenger travelling from Haryana roadways to different district. Read the data carefully and answer the questions. Q280. I. 𝑥 2 + 4𝑥 − 12 = 0 II. 𝑦 − 9𝑦 + 20 = 0 2 Q281. I. 𝑥 2 − 25𝑥 + 100 = 0 Total no. of passenger travelling from Haryana roadways = 22500 II. 𝑦 2 − 27𝑦 + 110 = 0 10% Q282. I. 𝑥 2 = 289 23% II. 𝑦 = √289 12% Q283. I. 𝑥 2 + 12𝑥 + 32 = 0 II 𝑦² + 7𝑦 + 12 =0 18% 22% Q284. I. 3𝑥 + 16𝑥 + 20 = 0 2 II. 𝑦² + 14𝑦 + 48 =0 15% Q285. I. 𝑥 2 + 𝑥 − 72 = 0 II. 𝑦 2 + 13𝑦 + 42=0 Gurgaon Hisar Sonipat Panipat Rewari Ambala Q291. No. of passenger who are travelling to Gurgaon are approximately how much percent less than no. of passenger travelling to Sonipat and Ambala together? (a) 75% (b) 78% (c) 50% (d) 65% (e) 90% Q286. I. 𝑥 2 + 5𝑥 + 6 = 0 II. 𝑦 2 − 9𝑦 + 14 = 0 Q287. I. 𝑥 2 − 14𝑥 + 45 = 0 II. 𝑦 2 + 2𝑦 − 35 = 0 Q288. I. 𝑥 2 + 11𝑥 + 18 = 0 Q292. What is the average no. of passengers who are travelling to Hisar, Panipat and Rewari? (a) 3025 (b) 2075 (c) 3375 (d) 3425 (e) 3075 II. 𝑦 2 + 6𝑦 + 8 = 0 Q289. I. 𝑥 2 + 5𝑥 = −6 II. 𝑦² – 15𝑦 = 16 Q290. I. 2𝑥 + 3𝑦 = 3 Q293. Passenger travelling to Hisar district are how many less than passenger travelling to Ambala? (a) 2525 (b) 2575 (c) 2425 (d) 2475 (e) None of these. II. 3𝑥 + 𝑦 = 8 Q294. If ratio of men to women who are travelling to Ambala and Gurgaon are 18:5 and 7:8 respectively, find ratio between men travelling to Gurgaon and women travelling to Ambala? (a) 5:7 (b) 7:18 (c) 7:5 (d) 14:15 (e) 15:8 33 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q295. If fair of a ticket for Rewari is Rs.75 and fair for 1 Panipat is 33 3 % more than that of Rewari, find difference between total revenue generated from both district (in Rs.)? (a) 33750 (b) 22025 (c) 34250 (d) 35750 (e) 25075 Directions (296-300): Paragraph given below gives information of literate and illiterate population out of total population of three cities i.e. A, B and C. Read the paragraph carefully and answer the following questions. Total population of city A and B are 22000 and 16000 respectively. Total literate population of city B is 6000 which is 6.25% of total population of city C. Ratio of literate to illiterate population in city A and C is 5:6 and 2:1 respectively. 40% of literate population in each city is graduate. Q296. Literate population from city B are what percent of illiterate population of city A? (a) 100% (b) 75% (c) 50% (d) 40% (e) 60% Q297. What is the ratio between graduate population of city C and total population of city B? (a) 5:8 (b) 3:5 (c) 5:3 (d) 8:5 (e) 1:3 Q300. If ratio of male to female in graduate population from city C is 9:7, find difference between graduate male from city C to literate but ungraduated from city B? (a) 7200 (b) 14400 (c) 10800 (d) 12000 (e) 11800 Directions (301-310): Bar graph given below shows quantity of five different products (i.e. rice, pulse, wheat, sugar and salt) sold (in kg) by a shopkeeper and table shows total revenue (in Rs.) generated by selling these individual products. Total amount sold of each product (kg) 60 55 50 45 40 35 rice pulses wheat sugar Name of product Total revenue (in Rs.) Rice 2200 Pulse 3750 Wheat 900 Sugar 1200 Salt 600 salt Q298. What is the difference between graduate population of city B and illiterate population of city C? (a) 29600 (b) 28400 (c) 28600 (d) 29400 (e) None of these. Q301. Cost price of per kg rice is how much more or less than per kg selling price of sugar when rice is sold at 60% profit? (a) Rs. 4 more (b) Rs. 5 less (c) None of these. (d) Rs. 4 less (e) Rs. 5 more Q299. Population which is literate but ungraduated from city A are what percent graduate population of city B? (a) 500% (b) 250% (c) 300% (d) 120% (e) 375% Q302. If 3 kg of wheat and 2 kg of salt is mixed, then what will be the selling price per kg of such mixture? (a) Rs.15 (b) Rs. 17 (c) Rs. 14 (d) Rs. 12 (e) Rs. 16 34 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q303. Total revenue generated from wheat is what percent of difference between total revenue generated from rice and salt? (a) 40.25% (b) 56.25% (c) 64.25% (d) 45.50% (e) 25.75% Q304. If cost price of per kg pulse is Rs. 60, find profit earned on selling 40 kg of pulse (in Rs.)? (a) 450 (b) 600 (c) 800 (d) 750 (e) 300 Q305. What is the average quantity of rice, pulse and wheat sold by shopkeeper? (a) 45 kg (b) 55 kg (c) 60 kg (d) 40 kg (e) 50 kg Directions (306-310): Study the following bar graph and answer the questions that follow. Given below is the bar graph which shows the number of students playing three different games in five colleges in year 2014. NOTE- one student plays only one sport Football Cricket Hockey 2 Q307. If 14 7 % of student playing Cricket of college N left playing cricket and started playing Football in same college then find the ratio of number of student playing football of college N and M together to the number of student playing Cricket of college K and N together? (a) 3 : 2 (b) 1 : 2 (c) 1 : 1 (d) 1 : 3 (e) 2 : 1 Q308. Average no. of students playing Hockey of college K, L and O is how much more than average number of students playing football of college K, L & M ? (a) 120 (b) 50 (c) 80 (d) 40 (e) 100 Q309. Total number of student playing Cricket of college L and M together are what percent more/less than total number of student playing Hockey of college K and M together? 1 (a) 32 3 % (b) 17 9 13 3 % 600 (c) 12 13 % 550 (d) 23 % 2 3 500 (e) 7 450 400 350 300 250 200 K L M N O 1 Q306. If 11 % of students playing Hockey of college L are 9 females then, number of males playing Hockey from same college is what percent of average number of students playing Hockey from college M & O ? 8 (a) 88 % 9 1 9 13 % Q310. If total number of students in college K in year 2015 is increased by 20% percent with respect to year 2014 and the ratio of student playing Football, Cricket and Hockey becomes 5 : 2 : 3 respectively then find the average number of students playing football in same college K in year 2014 and 2015 ? (a) 640 (b) 525 (c) 625 (d) 545 (e) 454 (d) 72 7 % Direction (311-315): Given bar graph shows total number of confirmed cases of COVIND-19 and number of deaths in four different countries. Study the bar graph carefully and answer the questions given below. (e) 82 3 % Mortality rate = 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒐𝒕𝒂𝒍 𝒄𝒐𝒏𝒇𝒊𝒓𝒎𝒆𝒅 𝒄𝒂𝒔𝒆𝒔 × 𝟏𝟎𝟎 (b) 63 % 3 8 (c) 68 9 % 2 2 35 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒅𝒆𝒂𝒕𝒉 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App 400000 Q315. If number of confirmed cases in China is increased by 25% and mortality rate remains same, what will be the new number of total deaths in China. (a) 4400 (b) 4500 (c) 4600 (d) 5200 (e) 5000 350000 350000 300000 250000 200000 140000 130000 80000 100000 50000 15000 17500 11000 4000 Spain Italy USA China 0 Total confirmed cases Deaths Q311. For which country mortality rate is lowest among the given four countries. (a) Italy (b) USA (c) Spain (d) China (e) USA and China Q312. Total confirmed cases in USA is what percent more than total deaths in Italy. (a) 1200% (b) 1350% (c) 2100% (d) 1900% (e) 1500% Q313. Find out the ratio between mortality rate of Spain to that of China? (a) 19: 11 (b) 43:14 (c) 15:7 (d) 14:9 (e) 13: 5 Q314. Total death in all four countries together is what percent of total confirmed cases in China? (a) 59.375% (b) 62% (c) 55% (d) 66.66% (e) 75% 36 www.bankersadda.com | Direction (316 – 320): Given below the bar graph shows the quantity of six different items (in kg) purchased by a person during the lockdown period. Read the data carefully and answer the questions. 35 Items purchased (in kg) 150000 30 30 25 25 20 20 18 15 15 12 10 Sugar Tea Salt Rice Oil Pulse Q316. If the sum of the price of one kg sugar and one kg salt is Rs.84 and the ratio of price of one kg of sugar and one kg of salt is 11: 10. Then, find the difference between the total price of Sugar and salt purchased by man? (a) Rs. 220 (b) Rs. 240 (c) Rs. 260 (d) Rs. 300 (e) Rs. 280 Q317. If the total price of tea is Rs. 900 and that of rice is Rs. 1500, then find the price of one kg tea is what percent more than that of rice? (a) 0% (b) 20% (c) 5% (d) 10% (e) 15% Q318. If the price of one kg of pulse and one kg of oil is Rs. 63 and Rs. 42 respectively, then find the ratio of the total price of the pulse to the total price of oil? (a) 13:25 (b) 1:2 (c) 3:5 (d) 18:25 (e) 12:13 www.sscadda.com | www.careerpower.in | Adda247 App Q319. The total quantity of sugar and salt purchased together by man is what percent of the total quantity of rice and pulse together purchased by man? 1 (a) 87 3 % 1 (b) 83 3 % (c) 74% (d) 92% Q324. Deepak takes 24 minutes more to cover a certain distance by decreasing his speed by 25%. What is the time taken by him to cover the distance with his original speed? (a) 70 minutes (b) 72 minutes (c) 75 minutes (d) 90 minutes (e) 84 minutes 1 (e) 64 3 % Q320. If the price of one kg salt, one kg rice, and one kg oil is Rs. 56, Rs. 32 and Rs. 40 respectively, then find out the total price of oil, salt, and rice purchased by man? (a) Rs. 2000 (b) Rs. 2800 (c) Rs. 2200 (d) Rs. 1800 (e) Rs. 2600 Q321. Train A crosses a 230m long platform in 29 seconds and train B crosses a 150m long platform in 24 seconds. Train B which is 450m long crosses train A in 160 seconds, while running in the same direction. Find how much time will the train A take to cross a 50m long bridge? (a) 16 seconds (b) 22 seconds (c) 20 seconds (d) 17 seconds (e) 25 seconds Q325. The speed of boat in downstream is ‘X-4’ kmph and ratio of time taken by a boat to cover a certain distance in upstream to downstream is 2 : 1. If boat takes 5 hours to Cover 40 km in Upstream, then find the value of X? (a) 16 (b) 20 (c) 22 (d) 24 (e) 18 Q326. Distance between two cities P and Q is 900 km. Car A and Car B can cover the distance between P and Q in ‘X’ hours and (X + 4) hours respectively. If Car B and Car A start from city P at 6.00 am and 8.00 am respectively and both Cars meet at 10.30 am, then find the distance between P and the point where both the cars meet? (a) 425 km (b) 475 km (c) 450 km (d) 500 km (e) 400 km Q322. A 950 metres long train-A crosses another train-B running in same direction in 16 seconds. If the ratio of speed of these trains is in the ratio 17:13 respectively, find out the length of train B? (a) 1000 meter (b) 1900 meter (c) 1600 meter (d) 1100 meter (e) Can’t be determine Q327. Downstream speed of a boat is 33 3 % more than its Q323. A train crosses a tunnel which is half of its length with a speed of 144 km/hr. in ½ min, then find the time in which it will cross another train which is double of its length and standing on platform in opposite direction with 60% of its initial speed ? (a) 120 sec. (b) 90 sec. (c) 150 sec. (d) 100 sec. (e) 180 sec. Q328. Amit goes to office from his home by bike at the speed of 30 kmph and he comes back to his home from office by bike at the speed of X kmph. If average speed for whole journey is 33 kmph, then find the value of ‘X’ (nearest to two decimal places)? (a) 35.56 km/hr (b) 36.00 km/hr (c) 36.67 km/hr (d) 32.50 km/hr (e) 34.50 km/hr 37 www.bankersadda.com | 1 upstream speed and the speed of the boat in still water is 15 km/h more than the speed of the stream. Find the total time taken by boat to travel 120 km in upstream? (a) 7 hr (b) 8 hr (c) 9 hr (d) 5 hr (e) 10 hr www.sscadda.com | www.careerpower.in | Adda247 App Q329. A train ‘X’ starts from station P at 8 am and reaches station Q at 4 pm. Another train ‘Y’ started from Q at the same time at which ‘X’ started and reaches ‘P’ at 3 pm. then find the time at which both the trains crossed each other. (a) 11 : 44 am (b) 11 : 48 am (c) 11 : 36 am (d) 12 : 44 pm (e) 11 : 50 am Q330. A car covered a certain distance at a certain speed in a fixed time. If car had moved 9 kmph slower, it would have taken 2 hours more and if it had moved 5 kmph faster, it would have taken 48 min less. Find the distance covered by car? (a) 300 km (b) 360 km (c) 320 km (d) 400 km (e) 450 km 1 Q331. Downstream speed of a boat is 57 % more than the 7 upstream speed of a boat. If the speed of the stream is 8 km/hr., then find the total time taken by the boat to cover 176 km in downstream and 70 km in upstream. (a) 7 hours (b) 6.5 hours (c) 7.5 hours (d) 6 hours (e) 8 hours Q332. Speed of a boat in still water is 8 km/h. It takes 5 hours to go upstream and 3 hours downstream distance between two points. What is the speed of stream? (a) 4 km/h (b) 2 km/h (c) 3 km/h (d) 1 km/h (e) 2.5 km/h Q334. A man can row 12 kmph in still water and it takes him 90 minutes to reach a place & return. If the speed of current is 4 kmph then how far is the place? (a) 8 km (b) 6 km (c) 10 km (d) 12 km (e) 16 km Q335. A man travels some journey on car with speed 60 kmph and some on cycle with speed 4 kmph. In return journey he come in train with speed 20 kmph and take equal time in both side journey. Find the ratio of the distance travel by car, cycle and train. (a) 8 : 2 : 11 (b) 3 : 2 : 5 (c) 2 : 1 : 3 (d) 6 : 1 : 7 (e) None of these Q336. A spherical ball of radius 16 cm is melted and casted into two cones of equal size and shape. If the base radius of the cone is 50% of the height of the cone. Find the height of each cone? (a) 36 cm (b) 18 cm (c) 32 cm (d) 20 cm (e) 16 cm Q337. How many three letters words starting with S (with or without meaning) can be formed out of the letters of the word, "STRANGE", if repetition of letters is not allowed? (a) 10 (b) 15 (c) 12 (d) 30 (e) 18 Q333. A man covers half of total distance with 12 km/h and another half distance with 24km/h. Find his average speed. (a) 12 km/h (b) 16 km/h (c) 10 km/h (d) 18 km/h (e) 6 km/h 38 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App Q338. If two dice are rolled simultaneously, find the probability of obtaining the sum (of numbers on these two dices) which is divisible by 2 or 3 but not by both? 1 (a) 4 1 (b) 2 (c) (d) 1 5 1 6 1 (e) 3 Q339. The area of a rectangular field having length 128m and breadth 16m is equal to the area of an isosceles right1 angle triangle. If the radius of a sphere is 12 % of the 2 hypotenuse of the isosceles right-angle triangle, then find out the total surface area of sphere? (a) 512π m2 (b) 343π m2 (c) 580π m2 (d) 494π m2 (e) 500π m2 Q340. Gurdeep Chhabra joined ‘Adda 247’ with the work experience of 26 years due to which average work experience of all employees of ‘Adda 247’ was increased by one year. If initial average work experience of all employees of ‘Adda 247’ was five years, then find the new number of employees in ‘Adda 247’? (a) 23 (b) 19 (c) 25 (d) 21 (e) 27 Q341. If two dices are rolled together, then find the probability of getting a number of one dice greater than the number on other dice? 3 (a) 4 2 (b) 3 (e) Q344. How many Words can be formed from the letters of the word ‘FLAGSHIP’ so that the vowels always come together? (a) 5040 (b) 10080 (c) 720 (d) 360 (e) 1440 Q345. One card is picked randomly from a pack of 52 playing cards. What is the probability that it would either be black queen or red king? 1 (a) 13 (b) (c) 5 13 6 13 7 (d) 13 (e) 8 13 Q346. The ratio of height of a cylinder to its base radius is 2:1 respectively. If radius of a hemisphere is equal to the radius of the cylinder, then find the total surface area of cylinder is what percent more than total surface area of a hemisphere? (a) 40% (b) 30% (c) can’t be determined 1 1 (d) 33 3 % 5 (e) 50% (c) 6 (d) Q343. How many cubes of 7.5 cm edge can be cut out from a cube of 45 cm edge? (a) 108 (b) 72 (c) 216 (d) 230 (e) 256 6 1 2 Q342. The radius of a cylinder & a sphere is same, and ratio of height and radius of cylinder is 2 : 1.If the volume of sphere is 288 π cm³ then find the volume of cylinder? (in cm³) (a) 438 π (b) 426 π (c) 420 π (d) 432 π (e) 444 π 39 www.bankersadda.com | Q347. A bag contains 4 red, 3 orange and 2 green color balls. Find the probability of selecting two same color balls from the bag? 1 (a) 2 (b) (c) 7 18 4 5 5 (d) 13 5 (e) 18 www.sscadda.com | www.careerpower.in | Adda247 App Q348. Find the probability of eight letters word that can be formed from the letters of the word ‘BLASTING’ so that vowels always come together. 1 (a) 4 (b) (c) 2 5 1 3 10 (d) 21 5 (e) 14 Q349. The total surface area of a cylindrical vessel is 1232 cm2 and the height of vessel is 2 times more than the radius of vessel. Find the volume of cylindrical vessel? (a) 4312 cm3 (b) 3201 cm3 (c) 3234 cm3 (d) 3256 cm3 (e) 3333 cm3 Q350. There are 5 red balls, 6 black balls and some green colored balls in a box. If the probability of choosing a black 1 ball from the box is 3, then find the number of greencolored ball in the box? (a) 5 (b) 4 (c) 6 (d) 8 (e) 7 Directions (351-355): Line graph below shows the number of complaints received by four different network operators (A, B, C & D) on two different days Tuesday & Wednesday. Study the line graph carefully and answer the following questions. 140 120 100 80 60 Q351. Find the difference between the total number of complaints received on both days by all network operators? (a) 40 (b) 50 (c) 60 (d) 70 (e) 80 Q352. Total number of complaints received by C & D together on Tuesday are what percent more/less than the number of complaints received by A & B together on Wednesday? (a) 62.50% (b) 63.63% (c) 66.66% (d) 33.33% (e)11.11% Q353. Find the ratio of number of complaints received by B on both days to number of complaints received by A & D together on Wednesday? (a) 1: 2 (b) 2 : 1 (c) 1 :1 (d) 5 : 4 (e) 4 : 5 Q354. Find the total number of complaints received by C on Tuesday and Wednesday are approximately what percent of total number of complaints received on Tuesday by all network operators together? (a) 36% (b) 60% (c) 53% (d) 48% (e) 67% Q355. Find the ratio of complaints received by A,B & D together on Tuesday to total complaints received on Wednesday by all network operators together? (a) 11: 17 (b) 21 : 31 (c) 18 : 19 (d) 29 : 32 (e) 51 : 43 Direction (356–360): In each of these questions a number series is given. In each series only one number is wrong. Find out the wrong number. 40 20 0 A B Tuesday 40 C D Wednesday www.bankersadda.com | Q356. 8, 4, 4, 10, 12, 30, 90 (a) 90 (b) 8 (c) 10 (d) 12 (e) 30 www.sscadda.com | www.careerpower.in | Adda247 App Q357. 11, 16, 25, (a) 41 (b) 66 (c) 11 (d) 151 (e) 25 41, 66, 102, 151 Q364. 3167 − 2881 − 112 =? −√1681 (a) 316 (b) 416 (c) 286 (d) 326 (e) 206 Q365. 62.5% 𝑜𝑓 ? −(5)2 = 152 (a) 200 (b) 100 (c) 500 (d) 400 (e) 300 Q358. 21, 25, 20, 28, 19, 27, 18 (a) 18 (b) 27 (c) 19 (d) 25 (e) 20 Direction (366–370): What approximate value should come in the place of question (?) mark in the following questions. Q359. 20, 28, 40, 56, 76, 104, 128 (a) 104 (b) 128 (c) 56 (d) 28 (e) 40 3 Q366. 24.01% of 449.98 + ?2 = (16.01)2 – √63.93 (a) 8 (b) 12 (c) 10 (d) 9 (e) 14 Q360. 1, 2, 6, 20, 88, 445, 2676 (a) 2 (b) 6 (c) 88 (d) 2676 (e) 20 Direction (361–365): What will come in the place of question (?) mark in following the question: Q361. 36 ÷ 4 × 7 + 4 × 4.5 =?2 (a) 9 (b) 7 (c) 19 (d) 17 (e) 3 Q368. 4? + 79.98% of 980.03 = 1039.99 (a) 4 (b) 2 (c) 3 (d) 5 (e) None of these Q362. √1849 − √256 = √?− √144 (a) 1681 (b) 1600 (c) 1296 (d) 1446 (e) 1521 Q369. www.bankersadda.com 1512.01 (a) 64 (b) 32 (c) 48 (d) 36 (e) 42 Q363. 250% 𝑜𝑓 30 − 175% 𝑜𝑓 36 + 52 =? (a) 27 (b) 18 (c) 37 (d) 21 (e) 31 41 Q367. ? × (44.01 % of 750.01 + 110.01) = 87.99% of 2499.98 (a) 2 (b) 4 (c) 3 (d) 5 (e) 6 ? + 49.99% of 488 = 70.03% of 399.99 Q370. ?% of 639.98 + 40.03% of 279.99 = (19.99) 2 (a) 25 (b) 50 (c) 35 (d) 45 (e) 40 | www.sscadda.com | www.careerpower.in | Adda247 App Direction (371-375): Read the given information carefully and answer the following questions. Line graph shows production of three products in terms of percentage (out of total production in the year) in four different years. Food Products Dairy products Beverages 70 Q374. The difference between number of food products and dairy products produced in 2015 and 2018 together is 12000. Find the average of dairy products and beverages produced by company in 2017? (a) 30000 (b) 22500 (c) 20000 (d) 24000 (e) 25000 Q375. Find the total production in 2019 if there was an increase of 20% in production in 2019 as compared to previous year given that number of dairy products in 2015 was 18000? (a) 1,20,000 (b) 1,08,000 (c) 1,18,000 (d) 1,12,000 (e) None of these 60 50 40 30 20 10 0 2015 2016 2017 2018 1.Company produces three different products i.e. food, dairy and beverages. 2.Total production of the company was same in all years. Q371. In 2016, quantity of food products and dairy products produced is what percent more or less than that of beverages produced in year 2015 and 2016 together? 1 (a) 13 % 3 (b) 15% 2 (c) 16 3 % (d) 12.5% (e) 10% Q372. If total number of products produced in year 2018 was 1,50,000. Find the difference between number of food products produced in 2017 and number of dairy products produced in 2015 and 2016 together? (a) 12000 (b) 18000 (c) 12500 (d) 10000 (e) 15000 Directions (376-380): What comes at the place of question marks: Q376. 588, (a) 552 (b) 510 (c) 542 (d) 532 (e) 572 562 , 614, 536, 640, Q377. 27, (a) 912 (b) 892 (c) 922 (d) 932 (e) 802 52, 102 , 202, 402, ? Q378. 17, (a) 461 (b) 481 (c) 471 (d) 491 (e) 451 41, 91 , 171, 293, ? www.careerpower.in | ? Q373. Find the ratio of average of number of food products produced in 2015, 2017 and 2018 to total number of beverages produced in 2016 and 2017 together. (a) 3: 2 (b) 2: 3 (c) 3: 5 (d) 5: 3 (e) 3: 4 42 www.bankersadda.com | www.sscadda.com | Adda247 App Q379. 35, (a) 9.62 (b) 8.76 (c) 12.56 (d) 10.08 (e) 11.02 7, 42, 8.4, Q380. 24, 60 , (a) 812.75 (b) 843.75 (c) 792.75 (d) 875.75 (e) 896.75 90, 225, 50.4, Q383. Find out the ratio between selling price of refrigerator C in 2018 and the selling price of refrigerator A in 2017? (a) 20:21 (b) 18:25 (c) 19:25 (d) 23:27 (e) 16:25 ? 337.5, ? Direction (381-385): Line graph given below shows the selling prices (in rupees) of three types of Refrigerators (A, B & C) in four different years i.e. 2016, 2017, 2018, and 2019 for a shopkeeper 35000 30000 25000 20000 15000 10000 2016 2017 Refrigerators A 2018 2019 Refrigerators B Refrigerators C Q381. If a discount of 24% is given on refrigerator C sold in 2018 and ratio of MP to CP of C in 2018 is 5 : 3. then find the difference between the discount allowed and profit earned on C in 2018. (in rupees) (a) 5000 (b) 4000 (c) 2000 (d) 6000 (e) 3000 Q382. Find out average selling price of refrigerator A in all the given years. (in rupees) (a) 16500 (b) 22500 (c) 18500 (d) 19500 (e) 25500 43 www.bankersadda.com | Q384. In which year sum of selling price of all 3 type of the refrigerator was the lowest? (a) 2019 (b) 2016 (c) 2017 (d) 2018 (e) 2016 and 2018 Q385. selling price of refrigerator A in year 2018 is approx. what percent of selling price of refrigerator B in 2019? (a) 78% (b) 88% (c) 82% (d) 72% (e) 93% Direction (386–390): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give the answers accordingly. (a) if x>y (b) if x≥y (c) if x<y (d) if x ≤y (e) if x = y or no relation can be established between x and y. Q386. I. x² – 14x + 48 = 0 II. y² – 17y + 72 =0 Q387. I. x2 + 13x + 42 = 0 II . y2 + 15y + 56 = 0 Q388. I . x2 + 8x + 12 = 0 II . 6y2 + 13y + 6 = 0 Q389. I. 2x2 + 9x + 9 = 0 II. y2 + 28y + 192 = 0 Q390. I . x² – 9x + 20 = 0 II. y² – 6y + 9 = 0 www.sscadda.com | www.careerpower.in | Adda247 App Directions (391-395): What should come in place of the question mark (?) in the following number series. Q391. 90, (a) 285 (b) 325 (c) 470 (d) 855 (e) 270 55, 75, 142.5, ? , 862.5 Q392. 5, (a) 850 (b) 750 (c) 800 (d) 805 (e) 820 12, 39, 160, 4836 Q393. 26, (a) 146 (b) 133 (c) 201 (d) 134 (e) 156 36, 54, 80, 114, ? ?, Q397. If the total price of tea is Rs. 900 and that of rice is Rs. 1500, then find the price of one kg tea is what percent more than that of rice? (a) 0% (b) 20% (c) 5% (d) 10% (e) 15% Q398. If the price of one kg of pulse and one kg of oil is Rs. 63 and Rs. 42 respectively, then find the ratio of the total price of the pulse to the total price of oil? (a) 13:25 (b) 1:2 (c) 3:5 (d) 18:25 (e) 12:13 Q394. 17, 25, 49, 97, 177, ? (a) 297 (b) 247 (c) 358 (d) 292 (e) 279 Q395. 21, 28, 42, 64, 95, ? (a) 125 (b) 158 (c) 142 (d) 136 (e) 164 Q399. The total quantity of sugar and salt purchased together by man is what percent of the total quantity of rice and pulse together purchased by man? 1 (a) 87 3 % Items purchased (in kg) Direction (396–400): Given below the bar graph shows the quantity of six different items (in kg) purchased by a person during the lockdown period. Read the data carefully and answer the questions. 35 30 25 20 15 10 Sugar 44 Tea Salt Q396. If the sum of the price of one kg sugar and one kg salt together is Rs.84 and the ratio of price of one kg of sugar and one kg of salt is 11: 10. Then, find the difference between the total price of Sugar and salt purchased by man? (a) Rs. 220 (b) Rs. 240 (c) Rs. 260 (d) Rs. 300 (e) Rs. 280 Rice Oil www.bankersadda.com Pulse | 1 (b) 83 3 % (c) 74% (d) 92% 1 (e) 64 3 % Q400. If the price of one kg salt, one kg rice, and one kg oil is Rs. 56, Rs. 32 and Rs. 40 respectively, then find out the total price of oil, salt, and rice purchased by man? (a) Rs. 2000 (b) Rs. 2800 (c) Rs. 2200 (d) Rs. 1800 (e) Rs. 2600 www.sscadda.com | www.careerpower.in | Adda247 App Solutions S4. Ans.(d) 20 Sol. Total number of students in A = 200 × = 40 S1. Ans.(d) 20 Sol. Total students in A = 200 × 100 = 40 100 22.5 Total students in C = 200 × 100 = 45 Difference between girls and boys in A = 50 × 20 100 10 = 10 Difference between girls and boys in C = 50 × 100 = 5 Let total number of girls in A = x So, total number of boys in A = (40 – x) And let total number of girls in C = y Total number of boys in C = (45 – y) Given, (40 – x) − 𝑥 = 10 2x = 30 x = 15 Similarly, (45 – y) – y = 5 2y = 40 y = 20 25 25 25 Required probability = 40 × 45 = 72 ( 25−𝑎 𝐶2 + 15−2𝑎 𝐶2 ) 40−3𝑎 𝐶 2 = 11 18 a=4 Required difference = 2a – a = a = 4 S2. Ans.(b) Sol. All five students from each B & D chosen for representing their respective school in a debate competition should be boys and number of girls should be greater than number of boys in order to maximize the probability of girls in remaining students. 15 Total number of students in B = 200 × 100 = 30 25 Total number of students in D = 200 × 100 = 50 20 Difference between boys and girls in B = 50 × 100 = 10 20 Difference between boys and girls in D = 50 × 100 = 10 Let total number of boys in B = a So, total number of girls in B = (30 – a) Let total number of boys in D = b So, total number of girls in D = (50 – b) ATQ, (30 – a) – a = 10 a = 10 Also, (50 – b) – b = 10 b = 20 20 30 8 Required probability = 25 × 45 = 15 100 S5. Ans.(a) Sol. Let total saving of Ritu be x Rs. 5𝑥 Invested in Scheme A = 15 𝑥 = 3 Rs. 4𝑥 Amount Invested in scheme B = 15 Rs. 2𝑥 Invested in scheme C = 5 𝑅𝑠. 10×10 Two year C.I on 10% =10+10+ 100 = 21% 15×15 Two year CI on 15% =15+15 + 100 = 32.25 % 20×20 Two year CI on 20% =20+20+ 100 = 44% ATQ, 4𝑥 32.25 𝑥 21 × 100 − 3 × 100 = 744 15 129𝑥 21𝑥 − 300 = 744 24x = 744 × 1500 744×1500 x = 24 x = 46500 Rs. Required difference 46500×6 44 46500×4 32.25 = × − × 1500 15 100 = 8184 – 3999 = 4185 Rs. S3. Ans.(a) 17.5 Sol. Total students in E = 200 × = 35 30 Difference between girls and boys in E = 50 × 100 = 15 Let number of girls in E = p So, number of boys in E = (35 – p) Given, (35 – p) – p = 15 p = 10 Number of boys in F = (35 – 10) – 5 = 20 And number of girls in F = 10 20×10 40 Required probability = 30𝐶 = 87 www.bankersadda.com 15 100 S6. Ans.(d) Sol. Let total students participated in exam P & Q be ‘x’ & ’y’ respectively. ATQ – 40 75 x × 100 × 100 = 900 3𝑥 = 900 x = 3000 40 (100−36) Also, y × × = 640 100 100 y = 2500 2500 Required ratio = =5:6 10 3000 2 45 20 Difference between girls and boys in A = 50 × 100 = 10 Let total number of girls in A = x So, total number of boys in A = (40 – x) ATQ, (40 – x) − 𝑥 = 10 2x = 30 x = 15 Let total number of boys who left school A = a And let, total number of girls left school A = 2a Now, | www.sscadda.com | www.careerpower.in | Adda247 App S7. Ans.(e) Sol. Let total students participated in exam from Q = a ATQ – 60 60 a× × = 1440 100 100 a = 4000 40 40 Total girls failed from Q = 4000 × 100 × 100 = 640 100 100 = 90 km/hr Car A speed in km/hr 750 18 = × 100 60 S8. Ans.(a) Sol. Let total number of students participated from S = p ATQ – 45 50 p× × = 1125 100 100 p = 5000 Given, ratio of total failed boys to total failed girls = 7 : 3 55 7 3 Required difference = 5000 × 100 × (10 − 10) = 1100 S9. Ans.(b) Sol. Let total students participated in exam from P = x 40 75 25 x × 100 × (100 − 100) = 600 𝑥 5 = 600 x = 3000 Let total students participated in exam from S = y 45 50 y × 100 × 100 = 1350 300 3300 −3000 3000 S13. Ans.(d) 1800 18 Sol. Speed of car B = 60 × 5 = 108 km⁄hr 1500 18 Speed of car C = 60 × 5 = 90 km⁄hr Average speed of Rajeev whole journey 2xy 2×108×90 = x+y = 108+90 19440 Required distance = speed × time 19440 = 198 × 11 = 1080 km Distance between city and village 1080 = = 540 km 2 × 100 = 3000 × 100 = 10% S10. Ans.(c) Sol. Let total students participated from P be ‘2x’ So, total students participated from R = 3x 55 (100−30) 40 (100−25) 3x × 100 × 100 – 2x × 100 × 100 = 2220 1.155x – 0.60x = 2220 ⇒ x = 4000 40 25 Total girls passed from P & R = 8000 × 100 × 100 + 12000 × 55 5 = 45 km/hr Let both Car has travelled for t hours before they meet. According to question 90t – 45t = 180 180 t= 45 = 4 hours. Therefore, total distance traveled is (90 × 4) + (45 × 4) = 540𝑘𝑚. = 198 km⁄hr y = 6000 55 Total failed students from S = 6000 × 100 = 3300 Required percentage = 1080 D= 5 𝑘𝑚 D=216KM 2D = 432 km 18 b = 2000 45 42 Total girls failed from R = 2000 × × = 378 100 D + 2 = 180 = 25 × 5 100 Required difference = 640 – 378 = 262 3 S12. Ans.(a) Sol. Car C speed in km/hr 1500 18 = 60 × 5 Let total students participated in exam from R = b So, 45 58 55 30 b× × −𝑏× × = 192 100 D 30 × 100 100 = 800 + 1980 = 2780 2780 Required average = 2 = 1390 S14. Ans.(c) 1200 18 Sol. Car E usual speed = 60 × 5 = 72 km⁄hr Time taken from this speed is 480 minutes, i.e. 8 hours. Hence total distance b/w City X & Y is 8 × 72 = 576 𝑘𝑚 Reduce speed = 72 – 12 = 60 km/hr 576 48 Required time = 60 = 5 hours or 576 minutes S15. Ans.(a) Sol. Time = t =time after which they meet 1200 18 Speed of car E = 60 × 5 = 72 km⁄hr Speed of car C = 1500 60 × 18 5 = 90 km⁄hr 2 = 72 (t – 3) + 90t = 762 S11. Ans.(b) Sol. Usual speed of car D in km/hr 900 18 = 60 × 5 = 15×18 5 = 72t – 48 + 90t = 762 762+48 =t= 162 = 54 km⁄hr Let total distance b/w Delhi and Lucknow is 2D km` According to question D D + = 10 54 t = 5 hr Both cars meet at = 3.20 + 5 = 8.20 pm 2 Distance from Delhi = (5 – 3) 72 = 312 km 36 46 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S16. Ans.(e) Sol. Pattern – 16 multiplied by consecutive prime numbers 16 × 2 = 32 16 × 3= 48 16 × 5 = 80 16 × 7 = 112 16 × 11 = 176 16 × 13 = 208 16 × 17 = 272 So, series is right no need to replace any number S17. Ans.(b) Sol. Wrong number = 8332 Pattern of series → 12 × 6 – 2 = 70 70 × 5 – 2 = 348 348 × 4 – 2 = 1390 1390 × 3 – 2 = 4168 4168 × 2 – 2 = 8334 8334 × 1 – 2 = 8332 So, 8332 should be replaced with 8334 S18. Ans.(c) Sol. Pattern of series I – 2.4 , 7.2 , 4, 12 , 18.4 , 26.4 , +3.2 +4.8 +1.6 +1.6 +1.6 +6.4 +1.6 +9.6 +8.0 +1.6 36 +1.6 Same series II – (A=18.4), 20 , + 1 .6 28 , (B=34.4), 42.4 23.2 , + 3 .2 + 4 .8 + 1 .6 + 1 .6 + 6 .4 + 1 .6 + 9 .6 + 8 .0 + 1 .6 + 1 .6 S19. Ans.(e) Sol. Pattern of series I – 123 = (112 +2) 171 = (132 + 2) 227 = (152 + 2) 291 = (172 + 2) 363 = (192 + 2) 443 = (212 + 2) 531 = (232 + 2) Same series II – 402 = (202 + 2) So, (A) = (162 + 2) = 258 (B) = (182 + 2) = 326 (C) = (242 + 2) = 578 47 (C=52) www.bankersadda.com S20. Ans.(a) Sol. Pattern of series I – Pattern of series I- Same series II – (A) = 1600 – 1280 = 320 (B) = 2400 + 560 = 2960 (C) = 3780 + 350 = 4130 S21. Ans.(c) Sol. Amar attempted total of 75 question in which 9 do not have negative marking. Total correct questions of Amar 80 = 100 × 75 = 60 If his all non-negative marking questions were correct, then his score = 60 × 3 – (15 × 1) – (0.5 × 25) [12.5 marks are subtracted for his un attempted questions] = 180 – 27.5 = 152.5 If his all non-negative marking questions were wrong, it means 6 of his questions which carry negative marks were wrong. = 60 × 3 – (6 × 1) – (0.5 × 25) = 180 – 18.5 = 161.5 S22. Ans.(d) Sol. Total questions attempted by Prem 80 = 100 × 75 = 60 Total correct questions of Prem 90 = 100 × 60 = 54 His 6 questions were wrong and 3 were negative marking questions. Therefore his score, is = 54 × 3 – (3 × 1) – (40 × 0.5) = 162 – 3 – 20 = 139 S23. Ans.(b) Sol. Prem, attempted total of 60 questions and 54 were incorrect. 34 questions of section A were correct. 50 He attempted, 16 question [ × 32] of section C and 6 were 100 incorrect. Score from section C = 10 × 3 – (6 × 1) – (16 × 0.5) = 16 If 6 questions from section C were wrong, it means all the questions, which he attempted from section B were correct. He attempted total of (60 – 34 – 16) = 10 question from section B Score from section B = 10 × 3 – 24 (0.5) = 30 – 12 = 18 Required difference = |16 – 18| = 2 | www.sscadda.com | www.careerpower.in | Adda247 App S24. Ans.(a) Sol. Correct question for Amar 80 = × 75 = 60 S30. Ans.(c) 240 Sol. Cost price per kg of sugar = 15 = 16 4 Discount offered by wholesaler = 20 × 100 = 20% 100 Score of Amar = 60 × 3 – (15 × 1) – (25 × 0.5) = 180 – 27.5 = 152.5 Score of Prem = 54 × 3 – (40 × 0.5) = 162 – 20 = 142 Required difference = 10.5 Discount offered by retailer to customer = = 357 Profit %= × 100 = 48.75% 2 3 65 × 100 = 3120 Total females do not get loan from village P 2 35 10 = 7200 × 3 × 100 × 21 = 800 Total males do not get loan from village P 1 = 7200 × 3 − 800 = 1600 Total males applied for loan from village P = 3120 + 1600 = 4720 Total number of males get loan from village T 80 20 8 = 9600 × 100 × 100 × 11 = 960 Total males do not get loan from village T 31.25 = 9600 × 100 − 960 = 2040 S32. Ans.(e) 125 72 Sol. Total females do not get loan from S = 10000 × 100 × = Rs. 1500 24 × 100 = 66.66% 100 16 × 27 = 1024 Total females do not get loan from Q 60 3 Required percentage = = Rs. 270 To be in a situation of no loss –no gain, he must sell remaining 50% at Rs. 270. 270 Price per kg = = Rs. 27 10 S29. Ans.(d) 70 Sol. Cost price of pulses for the retailer = 18 × 35 × 100 = Rs. 441 85 Cost price of 2 kgs of Almond = 2 × 40 × 100 = Rs. 68 Total CP = 441 + 68 = Rs. 509 304 1024 −720 720 × 100 2 = 720 × 100 = 42 9 % S33. Ans.(d) Sol. Total females do not get loan from Q = 8000 × 3 × 5 25 100 3 × = 720 5 40 Total males do not get loan from Q = 8000 × 100 − 720 = 2480 Total females do not get loan from S 72 24 16 = 10000 × 100 × 100 × 27 = 1024 Total males do not get loan from S 130 Total SP = 18 × 35 × 100 = Rs. 819 28 = 10000 × 100 − 1024 = 1776 310 2480 × 100 = 509 × 100 ≈ 61% www.bankersadda.com 25 = 8000 × 100 × 100 × 5 = 720 S28. Ans.(c) 90 Sol. Cost price of Rice for Retailer = 20 × 15 × 100 48 240 Total males applied for loan from village T = 5280 + 2040 = 7320 Required difference = 7320 – 4720 = 2600 Price at which he sold all the cashew = 40 × 30 × 100 509 117 S31. Ans.(a) Sol. Total number of males get loan from village P = 7200 × 68.75 S27. Ans.(b) Sol. Cost price of cashew for retailer = Rs. 900 819−509 × 100 = Total females do not get loan from village T = Rs. 475 Profit = 475 – 450 = 25 Profit % = 240 68.75 Price at which he sold this amount of wheat to customer 95 = 20 × 25 × 100 900 357−240 = 9600 × 100 × 100 = 5280 S26. Ans.(a) Sol. Cost price of 20 kg of wheat for retailer 90 = 20 × 25 × 100 = Rs. 450 1500−900 × 20% = 15% Selling price of mixture = (6 + 15) × 20 × 100 S25. Ans.(e) Sol. Let he attempted x correct questions. & (75 – x) wrong questions. But (75 – x – 4) carries negative marking. Therefore His score 3x – (71 – x) – 25 × (0.5) = 108.5 = 3x – 71 + x = 121 4x = 192 x = 48 hence option (e) Profit % = 75 100 85 Required ratio = 1776 = 155 ∶ 111 | www.sscadda.com | www.careerpower.in | Adda247 App S34. Ans.(c) Sol. Total number of males who get loan from P 2 65 = 7200 × × = 3120 3 S39. Ans.(d) Sol. Wrong number = 19480 Pattern of series – 2.5 × 24 = 60 60 × 12 = 720 720 × 6 = 4320 4320 × 3 = 12960 12960 × 1.5 = 19440 19440 × 0.75 = 14580 So, there should be 19440 instead of 19480. 100 Total number of males who get loan from Q 60 3 = 8000 × × = 3600 100 4 Total number of males who get loan from R = 8800 × 9 11 × 82 100 = 5904 Total number of males who get loan from S 72 76 = 10000 × 100 × 100 = 5472 Total number of males who get loan from T 68.75 80 = 9600 × 100 × 100 = 5280 S40. Ans.(a) Sol. Wrong number = 200 Number of males(second highest) who get loan are from village S. Total females applied for loan from village S 72 24 43 = 10000 × 100 × 100 × 27 = 2752 So, there should be199 instead of 200. S35. Ans.(e) Sol. Total females who get loan from village R 9 18 = 8800 × 11 × 100 = 1296 S41. Ans.(b) Sol. If each works 2 days at a time alternately starting with A, the work is completed in exactly 10 days. ∴ A works for 6 days and B worked for 4 days. 6 4 +b=1 ………….(i) a Total females who do not get loan from village R 4 = 1296 × 9 = 576 Total males who do not get loan from village R = 8800 × 2 11 If B starts, the work is completed in 10.5 days. ∴ B works for 6 days and A worked for 4.5 days. 6 4.5 + a =1 ………….(ii) b − 576 = 1024 Required average = 1024+1296 2 = 1160 By solving (i) and (ii) a = 9 days And, b = 12 days Time taken by A and B working together to complete the work 1 1 =1 1=1 1 S36. Ans.(c) Sol. Wrong number = 558 + + a b 9 12 36 1 = 7 = 5 7 days So, there should be 550 instead of 558. S42. Ans.(a) P 22P–36 1 ⇒ P² – 204P + 1188 = 0 ⇒ P² – 198P – 6P + 1188 = 0 ⇒ P = 198 or 6 ⇒P=6 Required number of balls in bag B = 48. S38. Ans.(e) Sol. Wrong number = 1260 S43. Ans.(c) Sol. Let that distance be D km. And, speed of boat P in still water be 2x km/hr Speed of boat Q in still water be 3x km/hr Speed of stream be r km/hr ATQ 𝐷 3𝐷 = 5×(2𝑥−𝑟) 3𝑥−𝑟 So, there should be 1258 instead of 1260 www.bankersadda.com 1 ⇒ P2 +60P+756 = 12 So, there should be 5.5 instead of 4.5. 49 P+2 Sol. P+18 – 42+P = 12 S37. Ans.(a) Sol. Wrong number = 4.5 Or, 𝑥 = 2𝑟 ……..(i) | www.sscadda.com | www.careerpower.in | Adda247 App Also, (3𝑥 + 𝑟) × 5 − 5 × (2𝑥 + 𝑟) = 60 or, 𝑥 = 12 speed of stream= 6km/hr S47. Ans.(d) Sol. Let, the income, expenditures and saving of P, Q and R: P Q R 3 3 Income x + 8000 x x 4 4 Expenditure on Rent y y y 30 Required ratio=30 = 1: 1 Solutions (44-45) Man → 12 × 6 days Woman → 8 × 18 days Child → 10 × 18 days (4 men + 12 women + 20 children)’s 2 day work 4 = 2[ 12×6 = + 12 8×18 + 20 10×18 ] 2 1 Remaining work = 1 − 2 = 2 According to question — 1 2 36 12×6 z z + 1000 z + 2000 Savings 6t 7t 4t Now, Total amount spent by all the three on food = 27000 ⟹ z + z + 1000 + z + 2000 = 27000 ⟹ z = 8000 Monthly income of Q = Monthly income of P + 6000 3 ⟹ x = x + 8000 + 6000 4 ⟹ x = 56000 Savings of P 6 = 1 1 Expenditure on Food Savings of Q 7 3 x + 8000 − y − z 4 6 ⟹ x − y − z −1000 = 7 =𝑥 Putting the values of x and z 42000 − y 6 ⟹ 47000 − y = 7 𝑥=1 ⟹ y = 12000 Monthly rent of the apartment = 3y = Rs.36000 S44. Ans.(c) Sol. A → 10 days B → 20 days S48. Ans.(e) Sol. Let S.P. → 120x So, C.P. of 1st Article → 100x C.P. of 2nd Article → 96x Profit difference = 4x → 8 Rs. x = 2 Rs. Selling price = 240 Rs. 2 days’ work of (A + B) = 3 12 days’ work of (A + B) = 18 A does the remaining work (20 – 18 = 2 units) 2 Required time = 12 + 2 = 13 days. S49. Ans.(d) Sol. Interest earned from first scheme S45. Ans.(a) Sol. Soldier = 56 × 1 × 24 days Required time = 56×24 42 R 𝑅 Rate of interest for second scheme = 𝑅 [1 + 100] = Interest earned from second scheme = 32 days S46. Ans.(a) Sol. Let, the income, expenditures and saving of P, Q and R: P Q R 3 x + 8000 x 3 Expenditure on Rent y y y Expenditure on Food z z + 1000 z + 2000 Savings 6t 7t 4t Income 4 4 100 …(ii) S50. Ans.(e) 4 2 Sol. Probability of a hit by Sunny = 6 = 3 1 For a miss = 3 5 Savings of Q = 62 2 % of income of Q = 8 x 4 5 5 7 8 14 ∴ Savings of R = × x = 1 For a miss = 2 x 1 6 2 2 1 1 2 1 1 1 1 2 www.careerpower.in | Adda247 App Required probability = + × × + × × × × +. .. 3 3 2 3 3 2 3 2 3 It formed a G.P. 2 1 1 1 In which a = and r = × = Percent of R’s savings out of his monthly income x = 14 3 × 100 3 x 3 2 a 6 2 4 Required probability = 1–r = 31 = 5 13 = 47 21 % 50 3 Probability of a hit by Satish = = Savings of Q and R are in the ratio 7 : 4. 4 x 15000×2×(100R+R2 ) = = 300𝑅 + 3𝑅2 100×100 ATQ, 300𝑅 + 3𝑅 2 − 1.5𝑅2 − 300𝑅 = 600 1.5𝑅2 = 600 ⇒ R = 20% 100𝑅+𝑅 2 1 Now, 5 2 = 15000 [(1 + 100) – 1] = 1.5𝑅2 + 300𝑅…(i) 1– 6 www.bankersadda.com | www.sscadda.com | S51. Ans.(d) Sol. Probability of a girl being selected from a section = Total girls in the section Total students in the section Let the number of girls, number of boys and total number of students respectively: For section A: 2x, 3x and 5x. For section B: 4y, 5y and 9y. For section C: 5z, 4z and 9z. According to the question, Ratio of total number of students in the three sections: ⟹ 5x : 9y : 9z = 10 : 12 : 9 ⟹x:y:z=6:4:3 Let the values of x, y and z be 6k, 4k and 3k respectively. Total number of girls in all the three sections = 2x + 4y + 5z = 12k + 16k + 15k = 43k Total number of students in all the three sections = 5x + 9y + 9z = 30k + 36k + 27k = 93k Probability of a girl being selected from the students from all the three sections together Total girls in all sections 43k 43 = Total students in all sections = 93k = 93 S52. Ans.(c) Sol. According to the question, Number of girls in sections A = Number of boys in section C ⟹ 2x = 4z ⟹ x = 2z Number of boys in section A : Number of boys in section C = 3x : 4z = 6z : 4z = 3 : 2 S53. Ans.(a) Sol. Let total quantity of milk = 200x L And total quantity of water = 100x L Total milk in P and Q = (20% + 15%) 200x = 35 × 2x = 70x L Total water in P and Q = 35 × x 25 25 Total water in X = 35x + 100 × 100 × 100x = 35x + 6.25x = 41.25x L Let cost price of milk per liter be Rs.10 So, cost price of (70x + 41.25x) L of mixture = 70x × 10 = Rs.700x Selling price of (70x + 41.25x) L of mixture = 111.25x × 10 = Rs.1112.5x Required profit percentage = = 1112.5x−700x 700x S54. Ans.(c) 50 Sol. Milk in vessel A and C = 100 × 2x = x Water in vessel A and C = × 100 7 825 = 14 = 0.55x Ratio of milk and water in M = x : 0.55x = 20 : 11 According to question, ⇒ 20 ×62 31 11 55x− ×62+17 31 x− x−40 6 =5 6 ⇒ 55x−5 = 5 ⇒ 5x – 200 = 3.30x – 30 x = 100 Quantity of milk in vessel B = 20 100 × 2 × 100 = 40 L S55. Ans.(b) Sol. Let total milk in all 5 vessel = 200x And total water in all 5 vessel = 100x So, 65 Total milk in all vessel except C = 100 × 200x = 130x 55 Total water in all vessel except C = 100 × 100x = 55x And Ratio of milk and water in vessel C = 35 × 2x : 45x = 70x : 45x = 14 : 9 According to question, 14 ×115 23 9 55x+ ×115 23 130x+ 130x+70 55x+45 9 =4 9 =4 520x + 280 = 495x + 405 25x = 125 x=5 Total quantity of water in all five vessel = 100x = 500 L 5x 3 C → 35 = 7 or x = 3 13 = 5814% Or we can say that profit in due the quantity of water in the mixture. So we can directly write 41.25x 70x × 100 = 5814% www.bankersadda.com From A and C Blue =15 Black=12 Red=8 So, required probability = 13 51 ×x S56. Ans.(c) Sol. A → Let number of black and blue colored balls be 4x and 5x respectively. B → Let number of red colored balls be y So, y+4x=5x+5 412.5 % profit = 55 100 | www.sscadda.com | 8c1 ×12c1 ×15c1 35c3 www.careerpower.in = 288 1309 | Adda247 App S57. Ans.(e) Sol. A → Let the length and breadth of the rectangle be 4x m and 3x m respectively. 4𝑥+3𝑥 Side of that square= = 3.5𝑥 𝑚 2 B → Sum of the lengths of diagonals of the rectangle = (𝑆𝑖𝑑𝑒 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒) × 3 − 4 C → Area of a square = 784 𝑚2 Hence, the question can be answered by using any two of the three statements together. Hence, It can be solved either by A and B together or by A and C together S58. Ans.(a) Sol. Let the ten’s place and unit’s place digits of the number be x and y respectively. And, the number be 10x + y = z A → z+z2=25z ⟹ z = 24 B → (10y + x) – (10x + y) = 18 ⟹y–x=2 10𝑥+𝑦 4 C → 𝑥+𝑦 = 1 ⟹ y = 2x Hence, either A alone or B and C together are sufficient to answer the question. 1 St B — 𝑦 = 𝑥 ⟹ 𝑥 = 4𝑦 St. C — 4 36 𝑥−𝑦 =4 ⟹ 𝑥−𝑦 =9 So, any two of the three statements are sufficient to answer the question. S64. Ans.(d) Sol. St A — Lengths = 4𝑥, 5𝑥 St B — ratio of speed = 1: 2 St C — speed of Ist train = 36 km/hr From B and C Speed of second train = 72 𝑘𝑚/ℎ𝑟 As we don’t know the directions of their motion so relative speed can’t be determined S65. Ans.(d) √3 √3 B MP-CP=Rs 480 C MP = SP + 384 Hence, Any two of three statements are sufficient to answer the question. S60. Ans.(a) Sol. From A and B A : B :C 8100×10 : 10800×10 : 9 : 12 : Profit share of B= Rs 1800 1800 Profit share of A= × 9 = 𝑅𝑠 1350 12 A and B together only are sufficient to answer this question S61. Ans.(b) Sol. Sol. A Profit percent = 25% B Let CP = 𝑥, SP = 1.25𝑥 New CP = 𝑥 + 500 1.25𝑥−(𝑥+500) 100 Profit percentage = × 100 = 9 𝑥+500 𝑥 = 4000 Profit = 1000 Rs. C C.P. = 𝑥 S.P. = 0.85(𝑥 + 1000) 0.85𝑥+850−𝑥 75 × 100 = 25 − 4 𝑥 𝑥 = 4000 Profit = (5000 − 4000) = 1000 Rs. So A and either B or C are sufficient. www.bankersadda.com S63. Ans.(e) Sol. Let the speed of boat in still water and speed of stream be x and y respectively. 45 St A — 𝑥 + 𝑦 = 3 ⟹ 𝑥 + 𝑦 = 15 Sol. St. C — 4 𝑎2 = 16√3, from here side of the equilateral triangle and height can be calculated. 48 St. B — Side of triangle = 3×2 = 8 S59. Ans.(b) Sol. Let the cost price be Rs 100x Then Marked price=100𝑥 × 1.6 = 𝑅𝑠 160𝑥 A SP = 112% of CP 4 7 And SP= 160𝑥 × 5 × 8 = 𝑅𝑠 112𝑥 52 S62. Ans.(e) Sol. As we don’t know the time for which Rinku borrowed the amount, so the rate of interest can’t be determined ℎ= 2 𝑎 St. A — no conclusion So using either B or C alone we can find the height. S66. Ans.(b) Sol. Quantity I Let the number be 10x +y Acc. to question y=x+2 and (10x +y) (x+ y) = 144 (10x + x+ 2) (x+ x + 2) =144 (11x+ 2) (x +1) = 72 11x² + 13x + 2= 72 11x² + 13x – 70 = 0 11x² + 35x – 22x – 70 = 0 On solving x = 2 Number is 24 Quantity II > Quantity I S67. Ans.(b) Sol. Quantity I Let they meet after ‘n’ days Applying Arithmetic progression n n [2 × 15 + (n − 1)(−1)] + [20 + (n − 1)2] = 165 2 n 2 2 [30 − n + 1 + 20 + 2n − 2] = 165 n² + 49n – 330 = 0 n = –55, +6 so, they will meet in 6 days | www.sscadda.com | www.careerpower.in | Adda247 App Quantity II Let required no. of days Quantity II: Let first we arrange 4 men in 3! Ways, then 4 women can be arranged in 4 places in 4P4 ways = 3! × 4P4 = 144 = 144 ×8 = 1152 Quantity I= Quantity II Days Quantity II > Quantity I S71. Ans.(a) Sol. From I → X + Y + Z = 16 × 3 + 12 = 60 Let, Y's 𝑎𝑔𝑒 𝑏𝑒 ′𝑎′ years Then, X= a –d Z = a+ d ⇒ 3a = 60, a = 20 Y’s present age = 20 years (X + Z)’s present age = 60 – 20 = 40 years From II → Let x, y and z years be the respective age of X, Y and Z 𝑥+𝑧 =𝑦 2 x + z = 2y z–y=4 x + z = 2(z – 4) ⇒ x + z = 2z – 8 ⇒ x – z = -8 ⇒z–x=8 Statement I alone is sufficient to answer the question but statement II alone is not sufficient to answer the question. S68. Ans.(a) Sol. Quantity I: Let present age of Randy = x x−10 12 = 24 − 19 x – 10 = 5×12 x = 70 years Quantity II: Required average = = 14× 111 −2×42 4 12 777 −84 2 12 609 203 = 24 = 8 = 25.375 year Quantity I > Quantity II S69. Ans.(a) Sol. Quantity I: Let C.P of 100 gm = 100 Rs So, he purchases 120 gm in 100 Rs 105 And sell 90 gm in = 100 × 100 RS S72. Ans.(d) Sol. According to concept if ratio between speed of two men is a : b then their distinct meeting points are (b – a) From I → Speed of Prabhat is 24 km/hr more than speed of Rahul. By this we can’t conclude, will they meet at half of the distance or not. From II → Ratio between speed of Rahul to speed of Prabhat →1:3 Distinct meeting points → (3 – 1) = 2 From statement (II) we can conclude, that they will meet at half of the distance on a circular track Hence, Statement II alone is sufficient to answer the question but statement I alone is not sufficient to answer the question. So, % profit = = = S.P.−C.P. C.P. 105 100 − 90 120 100 120 21 5 − 18 6 5 6 × 100 × 100 × 100 = 21−15 18 5 6 × 100 36 = 90 × 100 = 40% profit Quantity II: 50% → 12 Rs So, 100→ 24 Rs So, 80% → 19.2 There will be 0% profit if the book were sold for Rs.4.8 more Quantity I > Quantity II S70. Ans.(e) Sol. Quantity I: Let first we arrange all 4 men in 4! Ways then we arrange 4 women in 4P4 ways at 4 places either left of the man or right of the man. = 4! × 4P4 + 4! × 4P4 = 2 × 576 = 1152 53 www.bankersadda.com | S73. Ans.(d) Sol. Let there are ‘x’ red and ‘y’ yellow balls Where, x + y = 7 𝑦𝐶 2 From I → 10𝐶2 = 15 ⇒ 𝑦×(𝑦−1) 90 2 = 2 15 y (y – 1) = 12 y=4 From II → 3×𝑥 1 10 𝐶 = 5 2 ⇒ 3×𝑥 45 = 1 5 x=3 y=4 Either statement I or statement II by itself is sufficient to answer the question. www.sscadda.com | www.careerpower.in | Adda247 App S74. Ans.(b) Sol. From I → Let, number of pens and pencils sold be 2x and 3x respectively while cost price of a pen and a pencil be 3y and 2y respectively. Total cost price =2x × 3y + 3x + 2y = 12xy 137.5 Total selling price = 12𝑥𝑦 × = 16.5𝑥𝑦 100 From II → Profit % on selling one pen and one pencil is 25% and 50% respectively. If overall profit % is 37.5% ⇒ Using allegation No. of day required to complete the remaining work 800 = 50 + 3x 800 50 + 3x = 10 x = 10 10 additional women are required to complete the remaining work in 10 days. Quantity II: Let the additional number of men required be y. There are 4 + y men and 10 women now. Per day work of 4 + y men and 10 woman = (4 + y) × 5 + 10 × 3 = 50 + 5y units 800 No. of day required to complete the remaining work = 50 + 5y 800 50 + 5y ≤8 y ≥ 10 At least 10 additional men are required to complete the remaining work in either 8 or less than 8 days. Quantity II ≥ Quantity I Total cost price of pen = Total cost price of pencil S75. Ans.(e) Sol. Quantity I: Let vessel A contains 3x litres milk and x litres water and initial quantity of mixture in vessel A be 4x litres. Half of the content of vessel A is first poured into vessel B, then content of vessel B is poured into vessel C and finally contents of vessel C is poured into vessel A. So, vessel A finally contains contents of all the three vessels. Final ratio of milk and water in vessel A: Quanity of milk in all three vessels Quanity of water in all three vessels 3x + 30 9 x + 20 9 ∴ (8M+14W)×x×7 7 ×360 12 = (6M+10W)×15×6 5 ×360 12 171 𝑥 = 13 2 = 13 13 2 ∴ (Quantity I > Quantity II) ⟹ x = 20 Initial quantity of mixture in vessel A = 4x = 80 litres Quantity I = Quantity II S78. Ans.(b) Sol. S76. Ans.(d) Sol. 20 men can complete the work in 12 days. So, 1 man can complete the same work in 240 days. Efficiency of 5 women = Efficiency of 3 men 5W = 3M Ratio of efficiencies: M 5 = 3 W Let, a man does 5 units and a woman does 3 units of work per day & total units of work are 1200 units. 8 days’ work of 4 men and 10 women = 8 × (4 × 5 + 10 × 3) = 400 units Remaining work = 1200 – 400 = 800 units Quantity I: Let the additional number of women required be x. There are 4 men and 10 + x women now. Per day work of 4 men and 10 + x woman = 4 × 5 + (10 + x) × 3 = 50 + 3x units www.bankersadda.com 𝑥 = 10 ∴ Quantity I → 10 + 5=15 days Given 2M = 3W Quantity II → 13 13 =4 = 4 54 S77. Ans.(a) Sol. A B C Time 𝑥 + 5 𝑥 𝑥 − 4 1 1 1 ∴ 𝑥+5 + 𝑥 = 𝑥−4 | Let P is faster than Q Then P covers 72 km distance in the same time as Q covers 48 km distance Ratio of the speed = 72 : 48 =3:2 48 ∴Speed of faster train i.e., P = 2 × 3 = 72 km/hr Quantity 1→ Difference between P and Q = 72 – 48 = 24 km/hr. Let speed of train = T km/hr Let speed of car = C km/hr 120 480 ∴ + = 8 ………..(i) 𝑇 200 𝐶 400 1 + 𝐶 = 8 3 ………(ii) 𝑇 On solving (i) and (ii) T = 60 km/hr ∴ Quantity I < Quantity II www.sscadda.com | www.careerpower.in | Adda247 App S79. Ans.(a) 455 → 216 × 455 = 22750 Rs. S82. Ans.(b) Sol. Let no. of questions attempted correctly by him = x No. of questions attempted wrongly by him = y And no. of questions unattempted = z Then, x + y + z = 100 …(i) 4x – 2y – z = 320 …(ii) From (i) & (ii), 6y + 5z = 80 80 –5𝑧 or, y = Quantity I → 22750 Rs. max. value of y = 1 Sol. 20% = 5 180 → 216 5×36 → 6×36 25×6 → 36×6 ⇒ 150 → 216] Installments 125 → 216 125 → 216 Total Principal = 455 ∴ 216 → 10800 10800 6 80–5×4 6 60 = 6 = 10 1 And, 5% = 20 S83. Ans.(d) Sol. Let, total work = 48 units And efficiency A, B and C be a, b and C units respectively Then, a+b=4 b+c=3 and, b=2 hence, c = 1 and a = 2 A and B have same efficiencies Work done on day one= 2+2=4 units Work done on day two= 1 unit Work done in two days = 5 units Work done in 18 days= 45 units 3 Remaining three units will be done by A and B in 4 days 20×441 → 21×441 8820 → 9261 400×21 → 441×21 ⇒ 8400 → 9261] Installments 8000 → 9261 8000 → 9261 Sum = 25220 Rs. ∴ 25220 → 25220 25220 9261 → 25220 × 9261 = 9261 Rs. 240 Quantity II = 100 × 9261 = 22226.4 Rs. ∴ Quantity I > Quantity II S80. Ans.(b) 3 Hence, total no. of days taken=184 days. 6 Sol. Quantity I → 2 = 3 S84. Ans.(b) Sol. Let the MP of a chair and a table be Rs.5x and Rs.8x respectively. And, the number of chairs and tables bought be 6y and 5y respectively. CP of a chair for Abhishek = (100 – 20) % of 5x = Rs.4x CP of a table for Abhishek = (100 – 25) % of 8x = Rs.6x Total CP for Abhishek = 4x × 6y + 6x × 5y = 24xy + 30xy = 54xy SP of a chair for Abhishek = (100 – 25) % of (100 + 50) % of 4x = 4.5x SP of a table for Abhishek = (100 – 20) % of (100 + 50) % of 6x = 7.2x 2 Number of chairs sold in bunch of four by Abhishek = 3 rd of 6y = 4y 1 So, number of table sold for free by Abhishek = th of 4y = y 4 Total SP for Abhishek = 4.5x × 6y + 7.2x × (5y – y) = 27xy + 28.8xy = 55.8xy 55.8xy − 54xy 1.8xy 1 Profit % = × 100 = 54xy × 100 = 3 3 % 54xy 1 And, 3 𝜋𝑟 2 ℎ = 16𝜋 ℎ=3 ∴ 𝑦 = √42 + 32 = √16 + 9 𝑦=5 ∴ Quantity I < Quantity II S81. Ans.(a) Sol. Quantity of impurity in initial mixture = 5% of 2kg = 0.1 kg And let x kg of pure chili powder is mixed ATQ 0.1 2+𝑥 4 = 100 𝑥 = 0.5 𝑘𝑔 0.1 Profit %=2+0.5 × 100 = 4% S85. Ans.(c) Sol. According to the question, MP of a table = 300 + MP of a chair ⟹ 8x = 300 + 5x ⟹ x = 100 Total CP for Abhishek = 108000 ⟹ 54xy = 108000 ⟹ 54 × 100 × y = 108000 ⟹ y = 20 Number of chairs purchased by Abhishek = 6y = 120 To earn 30%, let he marks-up by x%, Then (4 + 𝑥 + 4×𝑥 100 ) % = 30% 104𝑥 4 + 100 = 30 104𝑥 100 = 26 or, x = 25% 55 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S86. Ans.(b) Sol. Let, initially planned tons of ore to mine per day be ‘𝑥’ and planned no. of days be ‘9𝑛’ 9𝑛𝑥 = 1800 …………..(i) ATQ, 3𝑛(𝑥 − 20) + (6𝑛 − 1)(𝑥 + 20) = 1800 …………(ii) or, 3𝑛𝑥 − 60𝑛 + 6𝑛𝑥 + 120𝑛 − 𝑥 − 20 = 1800 or, 60𝑛 − 𝑥 − 20 = 0 ……………(iii) 200 putting𝑛 = from (i) in (iii), 𝑥 𝑥 2 + 20 − 12000 = 0 ⇒ (𝑥 + 120)(𝑥 − 100) = 0 ⇒ 𝑥 = 100 S87. Ans.(c) Sol. Tons of ore mined till one-third of planned no. of days = 6 × 100 = 600 Two days prior to the planned date means, the task has to be completed in 16 days. 1 Tons of ore needed to be mined per day after rd period = 1800−600 16−6 S90. Ans.(b) Sol. Let length of longer train and smaller train be 8x m and 7x m respectively And that of speed of longer and smaller train be 3y m/s and 4y m/s respectively ATQ— (8x+7x)×7 (3y + 4y) = 90 15x = 90y x = 6y …………… (i) And 7x+(7x−100) 4y = 16 64y = 14x – 100………………… (ii) From (i) and (ii) y= 5 and x=30 speed of longer train= 15m/s and length of longer train =8 × 30 = 240 𝑚 240+110 1 required time= = 23 𝑠𝑒𝑐 15 3 1200 = 10 = 120 Extra tons of ore per day = 120 – 100 = 20 S88. Ans.(a) Sol. Let the number of students in A, B and C be b, b and c respectively. Let the total marks obtained by the students of class A, B and C be x, y and z respectively. Total marks obtained by the students of class A and B, B and C and A and C = 2(Total marks obtained by the students of class A, B and C) = 2(60(b + b + c)) = 120(b + b + c) This also equals 52.5(b + b) + 70(b + c) + 60(b + c) ∴ 120(2𝑏 + 𝑐) = 235𝑏 + 130𝑐 b = 2c Total number of students of A, B and C = b + b + c = 50 2(2c) + c = 50 ⟹c = 10 S89. Ans.(e) Sol. Number of students in all of the three sections- 20+20+10 =50 3 S91. Ans.(c) Sol. From I, Side of square = 24 cm Radius of cone = Side of square – 3 r = 24 –3 r = 21 cm From II, 66 7 Radius of circle = 22 × 2 = 10.5 𝑐𝑚 Height of cone = radius of cone + 7 .5cm = 10.5 + 7.5 = 18 cm From I & II 1 22 Volume of cone =3 × 7 × 21 × 21 × 18 = 8316 cm3 So, From I and II we can determine the area of cone. S92. Ans.(d) Sol. From I, Let man invested Rs 100 137.5 So amount after three years = 100 × 100 = 𝑅𝑠. 137.5 Interest = 137.5 − 100 = 𝑅𝑠. 37.5 37.5 ×100 R = 100×3 R = 12.5% From II, Total interest = 13200 − 9600 = 3600 𝑅𝑠. 3600×100 R= 9600×3 R = 12.5% So, either from I or II we can determine rate of interest. S93. Ans.(d) Sol. Let length of two trains be 9x meters and 8x meters From I, (90 + 72) × 5 18 = (9𝑥+8𝑥)9 68 153x = 3060 x = 20 meters Required difference = 20 × 9 − 20 × 8 = 20 𝑚𝑒𝑡𝑒𝑟𝑠 56 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App From II, Let length of smaller train is L meters 5 𝐿 90 × 18 = 6.4 L = 160 meters 160 Length of larger train = × 9 = 180 𝑚𝑒𝑡𝑒𝑟𝑠 8 Required difference = 20 meters So, either I or II alone sufficient to give answer of question. S94. Ans.(b) Sol. Quantity I – Total number of balls in bag = (5 + 6 + a + b) = (11 + a + b) ATQ – 𝑎 1 = (11+𝑎+𝑏) 6 6a – a - b = 11 5a − 𝑏 = 11---------- (i) 𝑏 2 Also, = 11+𝑎+𝑏 9 9b = 22 + 2a + 2b − 2𝑎 + 7𝑏 = 22 ----------------- (ii) From (i) & (ii) a=3 b=4 5×3 Required probability = 18×17 ( 2!) = 20 240 𝑦 = 80 y = 3km/hr x = 3 × 3 = 9 km/hr 240 Required time = (9+3) = 20 hours 6𝑥 32 = 3 64x = 3840 x = 60 km/hr 1200 Required time = 60 = 20 ℎ𝑜𝑢𝑟𝑠 Quantity I = Quantity II 20 20x = 56 + 7x + 7y 13x − 7𝑦 = 56 ------------------ (i) 𝑦 1 Also, 8+𝑥+𝑦 = 4 4y = 8 + x + y 𝑥 + 3𝑦 = 8 ---------- (ii) From (i) & (ii) we get y=5 x=7 8×7 Required probability = 2!× 20×19 28 = 95 Quantity I > Quantity II S95. Ans.(e) Sol. Quantity I – 22 Volume of cylindrical vessel = 7 × 17.5 × 17.5 × 18 = 17325 cm3 80 Volume of milk = 17325 × = 13860 cm3 100 30 × 7 × 3 × ℎ = 13860 462 h = 21 h = 22 cm Quantity II – Let length of rectangle and side of square be 12x & 11x respectively 4× 11𝑥 − 2(12𝑥 + 18) = 4 44x – 24𝑥 = 4 + 36 x = 2 cm Side of square = 2 × 11 = 22 𝑐𝑚 Quantity I = Quantity II www.bankersadda.com 4𝑦 1440+2400 = 51 Quantity II – Total dice = 8 + x + y 𝑥 7 = 57 96+144 Quantity II – Let speed of car be x km/hr ATQ – 240 480×5 32 + 6𝑥 = 3 𝑥 5 8+𝑥+𝑦 S96. Ans.(e) Sol. Quantity I – Let still water speed of boat x km/hr and speed of current y km/hr ATQ— (x + y) = 2 (x – y) x = 3y ATQ— 96 72 + (3y–y) = 20 (3y+y) | S97. Ans.(e) Sol. From statement (A) It can be observed that he obtained Rs. 900 in 3 years which means Rs. 300 in 1 year and Rs. 300 is 10% of Rs. 3000. Therefore, A + 3000 < 5000 A < 2000 It can’t be solved further From statement (B) We can conclude that A + 4000 > 5000 A > 1000 From statement (C) A = either of 500, 1000, 1500…….. By combining all of the statement we will get A = 1000 or 1500 Hence we can’t tell exact value of A. S98. Ans.(e) Sol. Let us assume selling price of item A and B is A and B respectively, while cost price of item A and B is x and y respectively. We have to calculate value of B. From statement (A) 4A + B = 70 7x = 70 x = 10 and 4y + x = 70 y = 15. But we can’t calculate value of B www.sscadda.com | www.careerpower.in | Adda247 App From statement (B) We obtain 4 equations from this statement B=y+3 13x = 10 A B-A=5 y–x=5 here are 4 equations and 4 variables, by solving we can get exact value of B as Rs. 18. From statement (C) Now profit % of both the items is 3 : 2. Let profit % be 3m% and 2m% for item A and item B respectively. From only C we cannot solve the question. By using A and C:ATQ 4 × 10 × (100+3𝑚) 100 + 15 × (100+2𝑚) 100 = 70 M=10 So value of B = 18 After combining statement (A) and (C) we can solve the question. Hence question can be solved by (B) alone or by using (A) and (C). S99. Ans.(c) Sol. (A) a + b = –3 There are many cases possible a = – 4, b = 1 ⇒ |a × b| < 10 a = –7 b = 4 ⇒ |a × b| > 10 (B) a×b<0 Either a > 0 & b < 0 OR b>0&a<0 (C) a + 2.5b = 0 a b x 2 2 2k 3 2k 8k = = y 5k say. +2= 3 15k + 12 = 16k k = 12 Hence age of Monu = 2k = 24 years. Hence, we can answer from either two statements. 8×8 2 year CI on 8% = 8 + 8 + 100 Let a = – 5x & b = 2x Minimum possible integers, which satisfies the eqn. are When x = 1, a = –5 and b = 2 ⇒ |a × b| = 10 or when x = 2 a = –10, b = 4 |a × b| > 10 Hence we can answer question from statement (C) alone. S100. Ans.(d) Sol. From (A) Age of Sonu is 36 And their total age is 96 years. We can’t tell more. From (B) Let age of Monu = 2x, age of Jonu is 3x. Now their average age of all three could be or From (A) and (B) We can tell age of Monu, as he is 33 ½% younger than Jonu, and it is given in the question 2 of them have same age. From (A) and (C) Average age is 32 years. Total age is 96 years. Monu is youngest, And age of Sonu and Jonu is y i.e. 36 years. From (B) and (C) Let age of Monu is x years. Age of Sonu & Jonu is y years. = 8% Principle invested in UCO bank –5 3 OR x+y+y y+2= 3 S101. Ans.(a) Sol. Given, rate offered by IDBI : rate offered by UCO = 3 : 4 6 Rate offered by UCO bank = 3 × 4 =2 2x+3x+2x From (C) Let age of Monu is x And age of Sonu and Jonu is y ATQ, x+y x+y+y +2= 3 2 = 16.64% 29160 Principle = 116.64 × 100 = 25000 Rs 10000×2×6 Amount obtained from IDBI = + 10000 100 = 1200 + 10000 = 11200 Required difference = 25000 – 11200 = 13800 Rs. S102. Ans.(c) Sol. Rate = 10% According to question Principle invested in SBI = 26250 100+4×10 × 100 = 18750 Rs. Amounts obtained from Yes bank 3 years CI on 10% = 33.1% 133.1 = 20000 × 100 = 26620 Rs. 18750 Required% = 26620 × 100 2x+3x+3x = 70.435% 3 We can’t tell further. 58 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S103. Ans.(b) Sol. Principle invested in ICICI 20000 = 5 ×7 = 28000 Rs Time = (3 – 1) = 2 year 15×15 2 year CI on 15% = 15 + 15 + 100 = 32.25% 32.25 Required interest = 28000 × 100 = 9030 Rs. S107. Ans.(a) Sol. Let Laddu’s prepared on Friday be 100x Then Laddu’s sold on Friday be 80x Also Laddu’s prepared on Monday be 100y Then Laddu’s sold on Monday be 90y ATQ, 100𝑥+90𝑦 113 = 100𝑥+100𝑦 120 𝑥 S104. Ans.(d) Sol. Let principle invested in UCO is X Rs. and principle invested in ICICI is (x + 3000) Rs According to question 15×15 2 year CI on 15% = 15 + 15 + 100 = 32.25% 8×8 2 year CI on 8% = 8 + 8 + 100 = 16.64% 132.25(𝑋+3000) 16.64𝑥 − 100 = 32870 100 115.61X = 3287000 – 396750 115.61x = 2890250 2890250 X= 115.61 = 25000 𝑅𝑠. Principle invested in ICICI = 25000 + 3000 = 28000 Rs. S105. Ans.(c) 6 Sol. Rate offered by SBI = 3 × 5 = 10% 4 Time = 2 × 1 = 2 𝑦𝑒𝑎𝑟 Principle invested in SBI 26250 = 100+(10×4) × 100 S108. Ans.(c) Sol. Let Barfi prepared on Friday be = 100a Laddu prepared on Friday = 100a + 80 Let Barfi prepared on Monday = 100b ATQ, 80 [100𝑎 + 80] = 60b 100 100a + 80 = 75b 15b – 20a = 16 …(i) Also, 100b 10 = ⇒ 7b = 10a or 14b = 20a 100a 112 = 10000 × 100 = 11200 Rs. Required sum = 11200 + 18750 = 29950 Rs. S106. Ans.(e) Sol. Let he prepared 300 x kg Laddu on Saturday, then Laddu’s sold on Saturday = 225x Laddu’s prepared on Sunday = 500x Laddu’s sold on Sunday = 400x Barfi’s prepared on Saturday and Sunday each 500𝑥+300𝑥 = = 400x 2 Barfi’s sold on Saturday = 280x Barfi’s sold on Sunday = 320x ATQ, (400x + 225x) – (280x + 320x) = 100 kg ⇒ 25x = 100 x=4 therefore, his Laddu’s remained unsold on Sunday = 100x = 400 kg. www.bankersadda.com And 90y × 20 – 80x × 20 = 11040 1800y – 1600x = 11040 90y – 80x = 552 …(ii) On solving (i) and (ii) we will get y = 16.8 100y = 1680 kg. 7 From (i) b = 16 or a = 11.2 Therefore, Barfi prepared on Friday = 1120 kg = 18750 Rs. Amount obtained from IDBI 100+(2×6) = 10000 × 100 59 5 ⇒ 𝑦 = 7 …(i) | S109. Ans.(a) Sol. Let he prepared 500x kg of Laddu & 400x kg of Barfi on Saturday. Quantity of Barfi sold = 280x kg 120x Profit from Barfi = 280x × 10 – 0.8 × 10 = 2800x – 1500x = 1300x Quantity of Laddu sold = 375x 125x Profit from Laddu = 375x × 10 – 0.8 × 10 = 3750x – 1562.5x = 2187.5x Required profit % 1300x+2187.5x = × 100 900x×200 3487.5x = 1800x = 1.9375 ≈ 2%% S110. Ans.(b) Sol. Let he prepared 100a kg of Barfi on each day. The laddu’s prepared = 75a, 125a and 105a. Barfi sold on these 3 days = 70a + 80a + 60a = 210a Laddu sold on these 3 days 225 = a + 100a + 94.5a 4 = 56.25a + 100a + 94.5a = 250.75a 460.75 Required % = × 100 = 76.15%≈ 76% www.sscadda.com 605 | www.careerpower.in | Adda247 App S111. Ans.(a) Sol. Let, one day work of a man, woman and youngster be 𝑚, 𝑤 and 𝑦 units, ATQ, 2𝑚 = 3𝑤 = 4𝑦 One-day work of 14 men, 12 women and 12 youngsters = 14𝑚 + 12𝑤 + 12𝑦 2 2 = 14𝑚 + 12 × 𝑚 + 12 × 𝑚 3 4 = 14𝑚 + 8𝑚 + 6𝑚 = 28𝑚 Total work = 24 × 28 𝑚 units To finish it in 14 days, 24×28𝑚 = 48𝑚 units must be done daily. 14 Which means, 48𝑚 − 28𝑚 = 20𝑚 additional units are to be done. For this, 20 more men are required. S112. Ans.(c) 1 Sol. To finish it in 19 = 19.2 days, 24×28𝑚 19.2 Solution (116-120) Let, Number of farmers producing RICE = 5x 120 ⇒ Number of famrers producing CORN = 5𝑥 × = 6𝑥 100 Let, Number of farmers producing only RICE and only CORN = a 3 Farmers producing only WHEAT = 5𝑥 × 5 = 3𝑥 Let, Number of farmers producing only WHEAT and RICE = y ⇒ Number of farmers producing only WHEAT and CORN but not RICE = 2y But Number of farmers producing only RICE is half of number of farmers producing RICE. ⇒ 𝑎 = 2.5𝑥 Number of farmers producing CORN is 48% of total farmers = 6𝑥 ⇒ 100% = 12.5𝑥 = Total number of farmers Number of farmers producing RICE and CORN 12 = × 12.5𝑥 = 1.5𝑥 100 5 = 35 m units are to be done daily. Which means, 35𝑚 − 28𝑚 = 7𝑚 additional units are to be done daily. 2 1 7 Now, 1. 𝑤 + 1. 𝑦 = 3 𝑚 + 2 𝑚 = 6 𝑚 7𝑚 Daily work of a pair of women & youngster is 6 units Hence, 6 such pairs are needed. ⇒ 1.5𝑥 + 2.5𝑥 + 𝑦 + 3𝑥 + 2𝑦 + 2.5𝑥 = 12.5𝑥 ⇒𝑦=𝑥 Number of farmers producing only RICE and WHEAT but not CORN = 24 = y Put value of ‘y’ in above diagram. By this we got, S113. Ans.(c) Sol. Profit on one articles = 30 Profit on 20 articles = 30 × 20 ⇒ 600 600 → 300 1 → 0.5 M.P. → 160 → 160 × 0.5 ⇒ Rs.80 S114. Ans.(e) Sol. Let total quantity ⇒ 1000 He gives → 800 → for → 100 Cost price of → 800 → 80 30 Now initial discount = × 100 = 18.75% Reduced discount = S.P. = 160×85 5 = 15% = 136 100 (136–80) Profit = 160 18.75×4 80 × 100 = 70% S115. Ans.(b) Sol. Let cost price of 20 articles → 20 × 100 = 2000 Actual profit on 20 articles → 20 × 30 = 600 S.P. → 2600 2600 S.P. of each undamaged article ⇒ 20–7 = 200 Initial discount = 30 160 × 100 = 18.75% S116. Ans.(d) Sol. Number of farmers producing at least two types of grain = 24 + 16 + 20 + 48 = 108 Number of farmers producing at most one type of grain = 60 + 72 + 60 = 192 108 Required % = × 100 = 56.25% 192 S117. Ans.(a) Sol. Number of farmers producing WHEAT and CORN = 16 + 48 = 64 125 Number of farmers producing Sugarcane = 64 × 100 = 80 200 So M.P. of each article should be = 81.25 × 100 ≈ 246 Approximately markup% = 146% S118. Ans.(c) 160 1 Sol. Required % = × 100 = 53 % 300 60 www.bankersadda.com | www.sscadda.com | 3 www.careerpower.in | Adda247 App S119. Ans.(b) Sol. Number of farmers producing wither only RICE or only WHEAT = 60 + 72 = 132 S120. Ans.(e) Sol. Required ratio = (24+20+48) 16 92 23 = 16 = 4 S121. Ans.(d) Sol. Total markers sold by A = 12% × 15,000 = 1800 1800 X marker sold by A = 9 × 4 = 800 Y marker sold by A = Z marker sold by A = 1800 9 1800 9 × 3 = 600 × 2 = 400 Let C.P. of one marker = ‘x’ 140 60 S. P. of X marker = 100 × x × 100 = 0.84x 140 80 S. P. of Y marker = 100 × x × 100 = 1.12x S. P. of Z marker = 140 100 ×x× 90 100 = 1.26x Total C.P. = [800 + 600 + 400]x = 1800x Total S.P. = 800 × 0.84x + 600× 1.12x + 400×1.26x = 672x + 672x + 504x =1848x 1848x−1800x 48x Total Profit Percentage = × 100 = 1800x × 100 = 1800x Let, S.P. of each marker = 2x, 3x, 4x Total marker sold by C 18 = × 15000 = 2700 100 X, Y and Z marker sold by C = 9 : 7 : 9 = 972; 756; 972 Total S.P =972 × 2x + 756 × 3x + 972 × 4x = 8100x Total S.P. of Y marker 756×3x×48600 = = Rs. 13608 8100 x S124. Ans.(b) 15 Sol. Total markers sold by ‘B’= 100 × 15000 = 2250 X, Y and Z markers sold by B =3:4:3 = 675; 900; 675 Satish Veer X markers sold = 60% 40% = 405; 270 Y markers sold = 40% 60% = 360; 540 Let S.P. of each X and Y marker = x, y ATQ 405x + 360y = 8370 …(i) 270x + 540y = 9180 …(ii) By solving (i), and (ii) x = 10, y = 12 2 23% S122. Ans.(b) 21 Sol. Total markers sold by E = 100 × 15000 = 3150 X, Y and Z sold by E = 3 : 2 : 1 = 1575; 1050; 525 Let S.P. of each marker sold by E = x, 1.5x, 3x Total S.P. = x × 1575 + 1.5x × 1050 + 3x × 525 = 4725x = 47250 ⇒ x =10 S.P. of x, y, z = 10, 15, 30 20 Total marker sold by F = × 15000 = 3000 9 18 6 14 3 25 4 20 X type of Marker sold by C = 25 × 100 × 15000 = 972 X type of Marker sold by F = 12 × 100 × 15000 = 1000 E sold maximum number of X type of markers S126. Ans.(a) Sol. Total work = 10080 units (LCM of days taken by all) 10080 Efficiency of A = = 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 70 10080 Efficiency of C = Efficiency of D = 90 10080 32 = 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 = 315 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 7 New efficiency of A = 144 × 8 = 126 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 2 New efficiency of D = 315 × = 210 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 10080 3 Required time = (126+112+210) = 22.5 ℎ𝑜𝑢𝑟𝑠 S123. Ans.(c) Sol. Let, total C.P. = x ATQ 10 8 x × − [x × ] = 9000 S127. Ans.(c) Sol. Total work = 10080 units (LCM of days taken by all) 10080 Efficiency of A = = 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 9 Efficiency of B= 2 x = 9000 9 Efficiency of E = x = 40,500 Total S.P. of marks if C wants to earn 20% profit 120 = 40500 × = 48600 Efficiency of F = 70 10080 60 10080 48 10080 36 = 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 = 210 units/hour = 280 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 3 New efficiency of F= 280 × 4 = 210 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 100 www.bankersadda.com 15 X type of Marker sold by E = 6 × 100 × 15000 = 1575 X, Y and Z sold by F = 4 : 5 : 3 = 1000; 1250; 750 Total S.P. of markers sold by F = 10 × 1000 + 15 × 1250 + 30 × 750 = 10,000 + 18,750 + 22,500 = Rs.51250 61 3 X type of Marker sold by B = 10 × 100 × 15000 = 675 X type of Marker sold by D = 15 × 100 × 15000 = 840 100 9 S125. Ans.(e) 4 12 Sol. X type of Marker sold by A = 9 × 100 × 1500 = 800 | www.sscadda.com | www.careerpower.in | Adda247 App ATQ – (210+210 )(𝑦) 168(𝑦+1) Total work in 2 hours = 802 – 427 = 375 units 10080×2 Total required time = [ 375 is the total work done in 2 2 =1 1344 = 25 84 144 S130. Ans.(d) 10080 Sol. Efficiency of E = 48 = 210 units/hour Efficiency of B = 2 Efficiency of C = Efficiency of C = Efficiency of D = Efficiency of E = Efficiency of F = 60 10080 90 10080 32 10080 48 10080 36 1680 = 210 units/hour = 15 days = 280 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 S131. Ans.(b) Sol. Relative speed of train A and Man = 750 5 5 = × + 10 × 3 75 100 = 126 𝑢𝑛𝑖𝑡𝑠/hour 60 n = 4.5 ℎ𝑜𝑢𝑟𝑠 Total time = n + (n + 23.5) = (4.5 + 4.5 + 23.5 ) = 32.5ℎ𝑜𝑢𝑟𝑠 S129. Ans.(d) Sol. Total work = 10080 units (LCM of days taken by all) 10080 Efficiency of A = = 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 Efficiency of D = Efficiency of E = Efficiency of F = 60 10080 90 10080 32 10080 48 10080 36 = 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 = 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 18 50 18 = 12.5 + 18 275 = 18 m/s 9 275 Length of train A = 150 m Distance in 24 sec = 24 × 12.5 = 300 m Length of platform = 300 – 150 = 150 m S132. Ans.(c) Sol Let length of train C and train F be 3L and 5L respectively Relative speed of train C to train F 2000 18 1000 18 = 3×60 × 5 + 60 × 5 = 40 + 60 = 100 km/hr 5 500 = 100 × 18 = 18 m/s According to question 500 = 210 units/hour L = 50 m Length of train C and F C = 3 × 50 = 150 m F = 5 × 50 = 250 m 18 = 280 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 www.bankersadda.com 108 Distance in 911 sec = 18 × 11 = 150𝑚 = 315 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 First hour total work of A, B, E and F = (144 + 168 + 210 + 280) = 802 units In Second hour total task destroy by C & D = −(315 + 112) = − (427) 62 = 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 Required days = 112 Let C work for n hours and G & H work for (n + 23.5) hours ATQ – 𝑛 × 112 + (216 + 126)(𝑛 + 23.5) = 10080 112n + 342(n+23.5) = 10080 112n + 342n + 8037 = 10080 112n + 342n = 10080 −8037 454n = 2043 2043 n = 454 Efficiency of C = 90 = 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 = 315 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 H in one hour = 168 × Efficiency of B = 60 10080 = 112 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 = 168 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 G in one hour = 144 × 2 = 216 units/hour 70 10080 10080 Work done by E & B: E = 210 x 12 = 2520 B = 168 x 35 = 5880 Remaining work = 10080 – (2520 + 5880) = 10080 - 8400 = 1680 units S128. Ans.(a) Sol. Total work = 10080 units (LCM of days taken by all) 10080 Efficiency of A = 70 = 144 𝑢𝑛𝑖𝑡𝑠/ℎ𝑜𝑢𝑟 10080 19 = 53 25 hours y = 4 hour Total work = 420y + 168(y+1) = 420 x 4 + 168(4+1) = 420 × 4 + 168 × 5 = 2520 units 2520 1 A will complete = = 17 ℎ𝑜𝑢𝑟𝑠 Efficiency of B = 375 hours ] 672 ×2 = 25 420y = 336y + 336 420y – 336y = 336 84y = 336 336 y= 𝑇= = | 3𝐿+5𝐿 = 14.4 150+(150+50) 2000 3×60 350×3×60 2000 www.sscadda.com 63 = 2 𝑠𝑒𝑐 | www.careerpower.in | Adda247 App S133. Ans.(a) Sol. According to question 180+180 18 Speed of train B = 21.6 × 5 S136. Ans.(b) Sol. Let number of lectures taken by professor on Monday and Tuesday each of second week be ‘x’. ATQ, 3 1 5 2 [ × 8 × 2 + × 𝑥 × 2 + × 6 × 2 + × 9 × 2] × 4000 = 4 2 3 3 = 60 km/hr Speed of train E in km/hr 300+300 18 𝐸 𝑠𝑝𝑒𝑒𝑑 = 30 × 5 600 18 = 30 × 5 = 72 km/hr Relative speed of train E and B when running in same direction 5 12×5 = 72 − 60 × = 18 10 18 = 3 m/s Time taken by faster train to cross slower train = = 480×3 10 (180+300)×3 10 = 144𝑠𝑒𝑐 2,00,000 ⇒ 12 + x + 20 + 12 = 50 ⇒x=6 S137. Ans.(c) 40 Sol. Required amount = [9 × 2 × 4000 + 8 × 2 × 6000 + 60 10 × 2 × 9000] 2 = [72,000 + 96,000 + 1,80,000] 3 2 = [3,48,000] = 2,32,000 = Rs. 2.32 lakh 3 S138. Ans.(e) Sol. Let no. of lectures taken by professor on third week be ‘x’ ATQ, 45 1,21,500 = 6000 × [7 + 5 + 𝑥 + 8] S134. Ans.(a) Sol. Given, Speed (m/s) : Taken time = 8x : 5x 120+240 8𝑥 = 5𝑥 60 ⇒ 20 + x = 27 ⇒x=7 7–5 Required % = 5 × 100 2 40𝑥 = 360 360 𝑥 = √ 40 = 3 2 = 5 × 100 = 40% Speed of train D = 8 × 3 = 24 m/s Taken time = 5 × 3 = 15 sec 120+600 S139. Ans.(b) 5×6000×45 Sol. Required ratio = 10×9000×40 24 Required ratio = (200+600)×60 =5∶8 1000 S135. Ans.(b) Sol. Given, Speed of smaller train D = 54 km/hr Speed of train B = V km/hr Relative speed = (V – 54) km/hr To cross the man, who sits in smaller train D, train B have to cross its own length with relative speed 5 180 = (𝑉 − 54) × 18 = 24 V = 81 km/hr 5 Required speed = 81 × 18 =3:8 S140. Ans.(c) Sol. Let no. of lectures given on Monday of second week be ‘x’ [𝑥×4000×2+5×6000×2+6×9000×2]60 112.5 = [8×4000×2+7×6000×2+6×9000×2]45 100 9 4 [8000𝑥+1,68000] ⇒8=3 27 = 32 [2,56,000] 8000𝑥+1,68,000 2,56,000 ⇒ 8000x = 2,16,000 – 1,68,000 ⇒ 8000x = 48000 ⇒x=6 45 = 2 m/s S141. Ans.(d) Sol. Let cost price of article X on Monday and Thursday be 400x and 500x respectively. Selling price of article X on Thursday 175 60 = 500𝑥 × 100 × 100 = 525𝑥 ATQ, 525𝑥 − 500𝑥 = 75 75 ⇒ 𝑥 = = 3 Rs. 25 Mark price of article Y on Monday 180 5 = 400 × 3 × 100 × 4 = 2700 Rs. Selling price of article Y = 2700 × 65 100 = 1755 Rs. Required profit = 1755 − 400 × 3 = Rs 555 63 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S142. Ans.(b) Sol. Let cost price of article X and Y on Friday = 100x 150 90 Selling price of article X = 100𝑥 × × 100 Profit on selling article X on Friday 150 90 = 500𝑥 × 100 × 100 − 500𝑥= 175x 100 = 135x Selling price of article Y = 135x + 520 ATQ, 135x – 100x + 135x + 520 – 100x = 1430 ⇒ 70x = 910 ⇒ x = 13 150 7 Mark price of article Y on Friday = 100𝑥 × × = 210𝑥 100 = 210 × 13 = 2730 Required % = 2730−[135𝑥+520] 2730 455 5 × 100 2 = 2730 × 100 = 16 3 % S143. Ans.(e) Sol. Let cost price of article X and article Y on Wednesday be 300x and 400x 126 Mark price of article Y on Wednesday = 400𝑥 × + 1008 100 = 504𝑥 + 1008 165 14 Mark price of article Y also equals to = 300𝑥 × 100 × 11 = 630𝑥 ⇒ 630x = 504x + 1008 1008 ⇒ 𝑥 = 126 = 8 165 S150. Ans.(c) Sol. Maximum number of bikes produced = 750 , on Thursday. S144. Ans.(c) Sol. Let cost price of article X = 100x Cost price of article Y = 80x 150 S151. Ans.(b) Sol. ATQ, 4 Mark price of article Y on Tuesday = 100𝑥 × 100 × 3 = 200𝑥 Selling price of article Y on Tuesday = 200𝑥 × 120𝑥−80𝑥 80𝑥 40 60 100 = 120𝑥 × 100 = 80 × 100 = 50% S145. Ans.(d) Sol. Let cost price of article X on Monday, Tuesday, Wednesday, Thursday and Friday be 100x, 200x, 300x, 400x and 500x respectively. Profit on selling article X on Monday 180 80 = 100𝑥 × 100 × 100 − 100𝑥 = 44x Profit on selling article X on Tuesday 150 80 = 200𝑥 × 100 × 100 − 200𝑥= 40x 70 100 R 2 = 10,000 [1 + 100] ⇒ 320R = 10,000 + R² + 200R – 8000 ⇒ R² – 120R + 2000 = 0 ⇒ R² – 100R – 20R + 2000 = 0 ⇒ R (R – 100) – 20 (R – 100) = 0 ⇒ (R – 20) (R – 100) = 0 ⇒ R = 20%, 100% S152. Ans.(e) Sol. Let he buy ‘x’ eggs out of which ‘y’ are rotten. ATQ, Total C.P. = 10x 108 Total S.P. = 100 × 10𝑥 = 10.8x 150 1080x = 800y + 1500x – 1500y 700y = 420x 𝑥 700 5 ⇒ 𝑦 = 420 = 3 Profit on selling article X on Thursday 175 60 = 400𝑥 × × − 400𝑥 = 20 100 www.bankersadda.com 8000×4×R 10.8x = 10y × 100 + 10(x– y) × 100 = 300𝑥 × 100 × 100 − 300𝑥= 46.5x 64 8000 + 80 Profit on selling article X on Wednesday 100 S147. Ans.(a) Sol. Total number of bikes produced by Bajaj from Monday to Friday = 900 S149. Ans.(c) Sol. No. of bikes produced on Tuesday and Thursday is same i.e. 150 = Rs 2772 Selling price of article Y = 504x = 504 × 8 = 4032 Required difference = 4032 – 2772 = 1260 165 520 70 = 300 × 8 × 100 × 100 Profit % = Frida y 180 140 200 S148. Ans.(e) 1150 Sol. Required average = 5 = 230 70 Selling price of article X = 300𝑥 × 100 × 100 165 Total profit = 44𝑥 + 40𝑥 + 46.5𝑥 + 20𝑥 + 175𝑥 = 325.5x ATQ, 44x = 176 ⇒x=4 Total profit = 325.5x = 325.5 × 4 = Rs 1302 Sol. (146-150) Mond Tuesd Wednesd Thursd ay ay ay ay Hero 180 150 250 150 Bajaj 160 220 200 180 Hond 200 200 300 250 a 540 570 750 580 S146 Ans.(b) 540 Sol. 750 = 18 ∶ 25 | www.sscadda.com | www.careerpower.in | Adda247 App Let total eggs = 5a Rotten eggs = 3a 2a 2 Required probability = = Alternative, Use Allegation method 5a S155. Ans.(b) Sol. Let the total volume of tank is 100m. Let efficiency of pipe B is 6x/hour And that of pipe A is 8x/hour Water flowing in the tank when all three pipes were opened in an hour= 6x + 8x – 40 = 14x – 40 ltr. (This is 25% of volume of tank+ 10 liters) 13 13 Water filled in tank in initial hours is 14𝑥 × = 5 7 7 26 𝑥 𝑙𝑖𝑡𝑒𝑟𝑠 Now this 14x-40 + 26x = 40x – 40 is total volume of tank. 25 Now 25 % of this total volume is 100 × (40x – 40)= 10x – 10 2 Required Probability = 5 S153. Ans.(a) Sol. Area of hexagon = Area of 6 equilateral triangles = Area of 3 parallelograms = Area of 2 equilateral triangle + Area of 2 parallelograms 6 ⇒ Area of parallelogram = 3 × 4√3 = 8√3 Let, side of equilateral ∆ = a √3 2 a = 4√3 4 ⇒a=4 Area of parallelogram = a × h Where a = side of hexagon on equilateral triangle = 4 cm and h = height of parallelogram So, 8√3 = 4 × h ⇒ h = 2√3 cm S154. Ans.(e) Sol. Total 100% work in completed by all four. Let A, B, C and D worked (a – 2d), (a – d), (a)% and (a + d) % work. a – 2d + a – d + a + a + d = 100% 4a – 2d = 100% But a = 30% ⇒ d = 10% A, B, C and D completed 10%, 20%, 30% and 40% work. C completes 30% work in 9 days. C completes 100% work in 30 days. B completes 20% work in 3 days. B completes 100% work in 15 days. A completes 10% work in 2 days. A completes 100% work in 20 days. Let D alone can complete 100% work in ‘d’ days. 4 4 4 4 + 15 + 20 + d = 1 30 1 1 1 d 1 4 1 30 ⇒ = –[ + 1 15 + 1 20 S156. Ans.(b) Sol. Let A fills 8x/hour water. While B fills 6x/hour water When all pipes were opened together They will fill (8x + 6x – 40) liters/hour. 14x – 40 liters is 10 liters less than 25% of tank’s efficiency. It means 25% efficiency of the tank will be (14x- 30) liters. 100 (14𝑥 − 30) = 56x – 120. Total efficiency of tank will be 25 In 4 hours, pipe A will fill 8𝑥 × 4 = 32𝑥 𝑙𝑖𝑡𝑒𝑟𝑠. ATQ, 32x = 56x – 120 X = 5. Hence volume of tank is 32×5= 160 liters. S157. Ans.(d) Sol. 80 speed of boat in still water = (10 + 10 × 100) km⁄hr = 18 km⁄hr ATQ— 280 280 560 + (18+10)+s + (18–10)+s = 45 (18+10) 280 28+s 8 28+s 560 + 8+s = 35 16 + 8+s = 1 64 + 8s + 448 + 16s = 224 + 28s + 8s + s2 s² + 12x – 288 = 0 s = 12 km/hr S158. Ans.(c) Sol. Let investment of A, B, C and D is a, b, c and d respectively. A B C 𝐷 ] Now in firt year → a × 12 : b × 12 : c × 12 ⇒ d = 10 ⇒ d = 10 D can complete 100% work in 10 days. ⇒ D can complete 40% work in 4 days. 65 ATQ, 10x – 10 + 10 = 14x – 40 4x = 40 X = 10 liters Hence, volume of tank is 4x – 40 = 400 – 40 = 360 litres. 360 Hence, C will fill tank in 40 = 9 hours. www.bankersadda.com In 2nd year → 2a × 12 : 4b 3 × 12 : www.sscadda.com | 5 6c In 3rd year | 6c 5 www.careerpower.in | × 12 × 12 : 𝑑 × 12 Adda247 App A:B:C:D 4 6 ⇒ (a × 12 + 2a × 12) : (b × 12 + b × 12) : c × 12 + 2 c × 12 : d S161. Ans.(b) Sol. Difference between second year’s interests × 12 = 12000 (1 + 7 2 %) (6 4 %) − 12000 (5 5 %) 3 7𝑏 5 17 1 43 3a : 3 : 5 𝑐 : d = 12 : 14 : 17 : 8 ⇒a:b:c:d=4:6:5:8 Difference between B and C initial investment = 1150 Total Investment of A and D together 1150 = × 12 = 13800 1 S159. Ans.(a) Sol. 1 4 1 29 = 12000 (40) (16) − 12000 ( 5 %) = 806.25 – 696 = Rs.110.25 S162. Ans.(c) Sol. Loan of Rs.20480 is settled by paying Rs.27778 after five years. Simple interest is applicable for first three years while compound interest is applicable for next two years. 3 1 3 4 4 4 27778 = 20480 (1 + (8 % + 5 % + 4 %)) (1 + 1 x 7 2 %) (1 + 100) 3 1 x 4 2 100 27778 = 20480 (1 + (18 %)) (1 + 7 %) (1 + 19 43 ) x 27778 = 20480 (16) (40) (1 + 100) 1 For second work 11 Efficiency of A = 9 × 9 = 11 units/day x = 64% S163. Ans.(d) Sol. Let the amounts borrowed under plan B and C be 19x and 13x respectively. ∴ Ratio of interests 9 Efficiency of B = 11 × 11 = 9 units/day Work done by A = 11 × x = 11x units. Work done by B = 9 × (x + 4) = (9x + 36) units ATQ, (9𝑥+36)–11𝑥 = 20𝑥+36 1 1 ⇒ 828 – 46x = 20x + 36 ⇒ 66x = 792 ⇒ x = 12 Required time = ( 20×12+36 20 1 3 = 16𝑥 × ((1 + 7 2 %) (1 + 6 4 %) − 1) : 13𝑥 × (8 4 % + 23 1 5 4 %) 43 17 14 = 16𝑥 × ((40) (16) − 1) : 13𝑥 × (100) 4 ) = 13 days. 5 91 14 = 16𝑥 × 640 : 13𝑥 × 100 S160. Ans.(d) Sol. Let original marked price be Rs 100x. 225 Then, New marked price of that article =100𝑥 × 100 = =5:4 𝑅𝑠 225𝑥 Selling price of article 3 4 8 = 225𝑥 × × × For old plan C = 8 4 % + 5 4 % + 4 4 % = 18 4 % 4 5 3 9 85 8 8 S165. Ans.(a) Sol. Let the amount borrowed under plan A be Rs.x Effective rate of interests for three years: 2 2 2 For plan A = 8 % + 6 % + 3 % = 19% 5 − 96𝑥) = 255 3 4 3 3 5 5 5 Total interest = 19% of x + 18% of (30000 – x) ⟹ 5540 = 18% of 30000 + 1% of x ⟹ x = Rs.14000 x = 24 cost price = 96 × 24 = Rs 2304 www.bankersadda.com 3 3 For plan E = 7 % + 5 % + 4 % = 18% x = 255 66 × 100 3 8 875𝑥 3 18 4 2 875𝑥 24𝑥 − ( 3 4 (20−18 ) =6 % = 225𝑥 × 8 × 3 × 6 ATQ 3 % Increase in interest = % Increase in effective interest rate for three years = Rs 96x Profit in first case=120𝑥 − 96𝑥 = 24𝑥 2nd selling price = 𝑅𝑠 3 2 = 2 1 For new plan C = 3 × 6 3 % = 20% = Rs 120x C.P. of article 100 = 120x × 125 7 S164. Ans.(c) Sol. Effective rate of interests for three years: | www.sscadda.com | www.careerpower.in | Adda247 App S166. Ans.(c) Sol. Candidate of party XYZ obtained 25 = × 300000 100 = 75000 votes Therefore, Total votes polled in constituency B 100 = 25 × 70000 S169. Ans.(b) Sol. Votes scored by winner candidate in constituency E = S167. Ans.(a) Sol. Votes scored by party XYZ from C 37.5 = × 300000 = 112500 100 As party obtained 50% votes, therefore Total votes polled in constituency E = 2 × 112500 – 45000 = 225000 – 45000 = 180000 Votes scored by winner candidates from E 65 = 100 × 180000 = 117000 Votes scored by party from E 12.5 = 100 × 300000 = 37500 Required answer = 117000 – 37500 = 79500 = 12.5 100 × 300000 = 37500 They lost by 117000 – 37500 = 79500 votes Hence winner candidate from constituency A got = 30000 + 79500 = 109500 Hence votes polled in A 100 = 30 × 109500 ≅ 365000 S170. Ans.(d) Sol. (i) If Party XYZ wins from C, total votes polled in constituency is 37.5 = 2 × 100 × 30000 = 225000 Candidate of party XYZ got 75000 votes from constituency B. If these constituencies have equal votes, winner candidate gets 25 = 100 × 225000 S168. Ans.(a) Sol. Votes secured by party XYZ from A 10 = 100 × 300000 = 30000 Votes secured by party XYZ from D 15 = 100 × 30000 = 45000 If party wins from A, total votes from A 100 = 30 × 3000 = 100000 votes from D winner candidate got = 45000 + 15000 = 60000 votes 100 Total votes polled in D = 40 × 60000 = 150000 votes Difference = 150000 – 100000 = 50000 If party wins from D, total votes from D 100 = 45000 × 40 = 112500 & winner candidate from A got = 30000 + 15000 = 45000 votes Total votes polled in constituency A 10 = 30 × 45000 = 150000 Difference = 37500. But maximum difference is of 50000, hence option a is the answer. www.bankersadda.com × 180000 = 117000 Votes scored by party XYZ = 30000 Runner up candidate got = 75000 – 7000 = 68000 Now for maximum number of candidates, all of the other should get min. possible votes = Remaining votes = 300000 – [75000 + 68000] = 157000 Min. votes possible 12000 157000 Maximum candidates = 12000 ≈13 Hence total number of candidates is 13 + 2 = 15 67 65 100 | = 57750 Hence it is not possible. (ii) Party got, 37500 votes from E, which is 35% of total polled votes (If we assume only 2 candidates). Hence winner must have more than twice votes. It is also not possible. (iii) It is possible, and we can’t check it whether party lost or won. (iv) it is also possible, as we don’t have information on number of votes. S171. Ans.(a) Sol. Total surface area of the toy = C.S.A of cone + C.S.A of Hemisphere Let, slant height of cone πrℓ + 2πr² = 858 cm² πr(ℓ + 2r) = 858 cm² ℓ =25 cm height of cone = 24cm volume of the toy 1 2 = πr 2 h + πr 3 3 1 = 3 πr 3 2 (h + 2r) 2 = 1950 cm3 3 www.sscadda.com | www.careerpower.in | Adda247 App S172. Ans.(e) Sol. Height of cylinder = 12 cm 11 Height of toy is 4 𝑡ℎ 𝑜𝑓 the height of cylinder = 33 cm Edges of cube = 33 – 12 = 21 cm C.S.A of cylinder = 2πrh = 66 × h r = 10.5 cm Total surface area of toy = (6a² – πr²)+ 2πrh + πr² (–πr², area subtracted due to aligment) 22 = 6 × 21 × 21 + 2 × × 12 × 10.5= 3438 𝑐𝑚2 7 S173. Ans.(c) 21 Sol. Sphere radius = 2 21 3 litre taxi P travel 1km In 10 litre taxi P travel ‘x’ km Price of petrol is Rs. 60/litre 60𝑥 So, Amount paid for petrol = 10 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑀𝑖𝑙𝑒𝑎𝑔𝑒 × 60 10 3 4 r+ h ∶ h ∶ 3r 13 : 6 : 7 12 Distance travelled by taxi ‘Q’ = 350 km For taxi ‘R’ r × 60 + r × 5 = 2592 (r= distance travelled by R) 15 S174. Ans.(d) 1 Sol. Volume of cone = 3 πr 2 h 1 1 10 𝑥 ⇒ x = 240 km Distance travelled by taxi ‘P’ = 240 km Similarly For taxi ‘Q’ q × 60 + q × 5 = 3500 (q= distance travelled by Q) height of cylinder = 12 required ratio 4 4 = πr 3 + πr 2 h ∶ πr 2 h ∶ πr 3 3 In Total amount paid = Amount for petrol + 5 × Distance travelled x ∴ 𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑝𝑎𝑖𝑑 𝑡𝑜 𝑡𝑎𝑥𝑖 ‘P’ = × 60 + x × 5 = 2640 So, cylinder radius= 2 4 Required ratio = 2 : 7 Solution (176-180) Let taxi P travelled x km. So, In 1 litre taxi P travel 10 km Distance travelled by taxi ‘R’ = 288 km For taxi ‘S’ s × 60 + s × 5 = 4500 (s= distance travelled by S) 18 22 = 3 × 7 × 7 × 7 × 24 = 1232 cm2 Volume of cuboid = 24 × 10 × 25% of 24 = 1440 cm² Difference = 1440 – 1232 = 208 cm² 208 Required % = 1232 × 100 = 16.88% ≈ 17% Distance travelled by taxi ‘S’ = 540 km For taxi ‘T’ 𝑡 14 × 60 + t × 5 = 2925 (t= distance travelled by T) Distance travelled by taxi ‘T’ = 315 km S175. Ans.(c) Sol. S176. Ans.(c) Sol. Distance travelled by taxi ‘Q’ = 350 km If millage of Q decreases by 2 So, amount paid to Q = 350 12–2 × 60 + 350 × 5 = 3850 Distance travelled by taxi ‘S’ = 540 km If millage of S decreases by 2 So, amount paid to S 540 = (18–2) × 60 + 540 × 5 = 4725 Total surface area of cone = πr × slant height + πr² = πr (ℓ + r) = πr (√8r 2 + r 2 + r) = πr (4r) = 4πr² Total surface area of remaining part of cylinder = πr² + 2πrh + πrℓ = πr (r + 2h + ℓ) = πr (r + 10r + 3r) = 14 πr² 68 www.bankersadda.com | Increase in amount = (3850 + 4725) – (3500 + 4500) = 575 S177. Ans.(d) Sol. Distance travelled by Q = 350 km 350 Average speed of Q = 25 = 14 km/hr Distance travelled by taxi ‘P’ = 240 km Average speed of P = Required % = www.sscadda.com | 15–14 15 240 16 = 15 km/hr 20 2 × 100 = 3 % = 6 3 % www.careerpower.in | Adda247 App S178. Ans.(a) Sol. Distance travelled by S = 540 km 540 Average speed of S = 30 = 18 km/hr Let two speeds with which taxi S travel is 3x km/hr and 2x km/hr So, 2×3x×2x 3x+2x 12x 5 = 18 = 18 S182. Ans.(d) Sol. Let present age of Simmi and Rimmi be 4x yr and 3x yr respectively And let present age of Rina be y yr. ATQ 4𝑥+6 3 =1 𝑦 4𝑥 + 6 = 3𝑦……….. (i) And 3𝑥 2 = 𝑦 18×5 x = 12 = 7.5 270 Time for higher speed =3×7.5 = 12 hour Alternate, Speed is inversely proportional to time if distance is same So, Higher speed : Slower speed = 3 ∶ 2 Time in Higher Speed : Time in lower speed = 2 ∶ 3 Total time = 30 hours 𝐼𝑓 5 → 30 30 ⇒ 2 → 5 × 2 = 12 ℎ𝑜𝑢𝑟𝑠 1 ⇒3x=2y …………. (ii) From (i) and (ii) y= 18 yr present age of Simmi=48 yr and that of Rimmi= 36yr 18+48+36 required average = 3 = 34 𝑦𝑟 S183. Ans.(d) Sol. Let the value of x and y be 9a and 7a respectively 720×100 CP = (100+9a) Now, 720×100 SP = (100+9a) & CP = Rs. 720 ATQ, S179. Ans.(d) Sol. Distance travelled by taxi ‘R’ = 288 km Distance travelled by taxi ‘T’ = 315 km 720×100×100 (100+9a)×(100–7a) = 720 ⇒ 10000 = 10000 + 200a – 63a² ⇒ 63a² – 200a = 0 200 ⇒ a = 0 or a = 63 288 Average speed of R = 18 = 16 km/hr 315 Average speed of T = 15 = 21 km/hr Required SP = 720×(100+7× 100 Ratio = 16 : 21 𝑥×100 Breadth of second field = (100+4𝑎) cm ATQ, x×100 (100–a) 240+350+288+540+315 5 = 346.6 km S181. Ans.(e) Sol. Let initial quantity of milk and water be 9x lit and 2x lit respectively. ATQ 9𝑥−44× 2𝑥−44× 9 11 2 +12 11 3 =1 3 New quantity of milk=9 × 16 − 36 + 64 × 8 = 132𝑙𝑖𝑡 5 And that of water=2 × 16 − 8 + 12 + 64 × 8 = 76 𝑙𝑖𝑡 132 69 y×100 × (100+4a) = xy ⇒ 10000 = (100 – a) (100 + 4a) ⇒ 10000 = 10000 + 400a – 100a – 4a² ⇒ 4a² – 300a = 0 ⇒ a = 0 or 75. S185. Ans.(b) Sol. Let the amount of Aman and Vikash be Rs. 3x and 5x respectively. Total amounts = Rs. 8x Total amounts after 2 years 20 ⇒ 𝑥 = 16𝑙𝑖𝑡 Required ratio= 76 = 880 S184. Ans.(b) Sol. Let length and breadth of first rectangular field is x cm and y cm respectively Area of first rectangular field = xy cm² 𝑥×100 Length of second field = (100–𝑎) cm S180. Ans.(e) Sol. Distance travelled P → 240 km Q → 350 km R → 288 km S → 540 km T → 315 km Required Average = 200 ) 63 = 33: 19 2 20 = 3x × (1– 100) + 5x × (1 + 100) 4 4 6 6 5 5 5 5 2 = (3x × × ) + (5x × × ) = 1.92𝑥 + 7.2x = 9.12x Change = Rs. 9.12𝑥– 8x = Rs.1.12𝑥 1.12𝑥×100 % change = = 14% 8𝑥 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S186. Ans.(c) Sol. On Tuesday Gaurav = 25×100 Abhishek = Neeraj = = 50 minutes 50 20×100 80 10×100 100 S190. Ans.(a) Sol. Let Arunoday worked for 𝑥 minutes 2 5 5 5 𝑥 ∴ 20 + 25 + 35 + 26 + 250 = 1 𝑥 = 25 minutes 250 𝑥 ≈ 91 minutes ∴ Required time = 91 – 5 = 86 minutes = 10 minutes Clearly on Tuesday, the efficiency of Neeraj is maximum. So he should start the job so that the job is completed in the least possible time. S187. Ans.(b) Sol. On Tuesday Gaurav = 50 minutes Arunoday = 150×100 5 = 100 11 89 100 1 1 + 500 25 3𝐶 2 × 5 𝐶2 8𝐶 3 𝐶1 8𝐶 3 15 = 28 15 = 56 Quantity II: boys = 10; girls = 10 10×9×8×7 10×9×8×7×6 Required ways = 10𝐶4 + 10𝐶5 = 4×3×2×1 + 5×4×3×2×1 = 462 Quantity I>Quantity II = 125 min. Neeraj = 10 min. (Abhishek + Shailesh + Neeraj)’s 1 minute work = 10 + 2 + 25 = 37 units Shailesh will work on this job for 7 minutes. 7×2 ∴ Share of Shailesh = × 875 = 49 Rs. 250 S189. Ans.(e) Sol. On Tuesday — Aman = 125 × 2 = 250 min. Neeraj = 10 min. Abhishek = 25 min. Aman’s 50 min. work = 50 = 1 250 5 1 4 Remaining work = 1 − 5 = 5 4 let l and b S195. Ans.(e) Sol. Quantity I: Given a&b are roots of 𝑥 2 + 𝑥 − 6 = 0 Sum of roots, 𝑎 + 𝑏 = −1 Product of roots, 𝑎𝑏 = −6 1 1 𝑎+𝑏 −1 1 Required value = 𝑎 + 𝑏 = 𝑎𝑏 = −6 = 6 Quantity II: we know, (𝑎 + 𝑏)2 = 𝑎2 + 𝑏 2 + 2𝑎𝑏 = 20 + 2(8) = 36 𝑎 + 𝑏 = ±6 1 1 𝑎+𝑏 6 3 Required value = 𝑎 + 𝑏 = 𝑎𝑏 = 8 = 4 (𝑤ℎ𝑒𝑛 𝑎 + 𝑏 = +6) 1 + 1 𝑎+𝑏 −6 −3 No relation between quantity I & II. 10 25 www.bankersadda.com S194. Ans.(e) Sol. Quantity I: perimeter = 2(l+b) = 48cm; are 5x cm & 3x cm respectively. 8x = 24cm X = 3cm Therefore , Length= 15 cm and Breadth= 9 cm Area = l × b = 15×9= 135 cm2 Quantity II: 135 Quantity I=Quantity II Required value = 𝑎 + 𝑏 = 𝑎𝑏 = 8 = 4 (𝑤ℎ𝑒𝑛 𝑎 + 𝑏 = −6) 5 Required time = 1 5 1 = 5 7 minutes 70 Quantity II: required probability = 3𝐶 1 × 5 S193. Ans.(a) Sol. Quantity I: boys = 12; girls = 20 – 12 = 8 12×11×10×9 8×7×6 Required ways = 12𝐶4 + 8𝐶5 = 4×3×2×1 + 3×2×1 = 551 4 = 21 21 minutes S188. Ans.(d) Sol. On Tuesday — Abhishek = 25 min. 40 20 Quantity I>Quantity II 89 Remaining work = 1 − 100 = 100 50×100 × 60 = 100 𝑚𝑖𝑛 1 11 Shailesh = 120 Sol. Quantity I: required probability = (Gaurav + Arunoday)’s 5 minutes work = 50 + 500 = 10 + Required time = 200 S192. Ans.(a) 5 100 Required time = Quantity I<Quantity II Abhishek = 25 minutes 1 S191. Ans.(c) Sol. Quantity I: Prabhas travels a distance of 100 km in 1 hour when he stops for 10 min. Actual travel time = 60 – 10 = 50 min 100 Speed of car = 50 × 60 = 120 𝑘𝑚𝑝ℎ Quantity II: required time = 7+3 = 2 ℎ𝑜𝑢𝑟 = 120 𝑚𝑖𝑛 = 500 minutes 30 578 = 1 − 910 | www.sscadda.com | www.careerpower.in | Adda247 App S196. Ans.(c) Sol. From I – Let income of Sameer = 25x 96 So, income of Veer = 25𝑥 × 100 = 24𝑥 Let expenditure of Veer = 7y So, expenditure of Sameer = 8y 3 Deepak spend th of his income. S199. Ans.(b) Sol. Let radius of circle = r cm So, side of square = r + 3.5 cm From I – 22 2 × × 𝑟 − 2𝑟 = 45 7 5 From II – Saving of Sameer = 7000 Rs. Saving of Veer = 7400 And, Income of Deepak is Rs. 1000 more than that of Sameer From I & II – (25𝑥−7000) 8𝑦 = (24𝑥−7400) 7𝑦 r = 10.5 cm side of square = 10.5 + 3.5 = 14 cm Area of square = 196 cm2 Statement I alone is sufficient to give answer. From II – Let breadth of rectangle = 2x So, radius of circle will be = 3x ATQ – 22 ×3𝑥 7 17x = 10200 ⇒ x = 600 Rs. Income of Deepak = 25 × 600 + 1000 = 16000 𝑅𝑠. 2 Saving of Deepak = 5 × 16000 = 6400 Rs. 2× S197. Ans.(b) Sol. Let cost price = 100x Marked price = 140x From I – 75 140x × − 100𝑥 = 50 S200. Ans.(e) Sol. From I – Difference between speed of train A & B =10 meters/sec And, length of train B = 240 meters From I, we can’t determine From II – Train B cross pole in 8 sec And train B cross train A in 12 sec From II, we can’t determine From I & II – 240 Speed of train B = 8 = 30 𝑚𝑒𝑡𝑒𝑟𝑠/𝑠𝑒𝑐 2(2𝑥+15) Respective ratio of saving of Veer & Deepak = 7400 : 6400 = 37 : 32 So, Statement I & II together is sufficient to give answer of the question. 100 5x = 50 ⇒ x = 10 Rs. Cost price = 1000 Rs. Statement I alone is sufficient. From II – 6 90 (140x × 7 × 100 ) − 100𝑥 = 80 8x = 80 ⇒ x = 10 Rs. Cost price = 1000 Rs. Statement II alone is sufficient. So, either statement I or II alone is sufficient to give answer of the question. (7−𝑥)(6−𝑥) x = 3.5 cm Radius of circle = 10.5 cm side of square = 10.5 + 3.5 = 14 cm Area of square = 196 cm2 So, either statement I or Statement II alone is sufficient. Speed of train A = 30 − 10 = 20 meters/sec Let length of train A = L meters 240+𝐿 So, (30 + 20) = 12 L = 600 − 240 L = 360 meters So, Statement I and II both together sufficient to give answer of the questions S198. Ans.(a) Sol. Given, number of green balls = 5 So, let total number of blue balls = x So, number of red balls = (7 – x) From I – 𝑥 7−𝑥 7 + 12 = 12 12 So, we can’t determine value of x from statement I From II – 𝑥 (𝑥−1) 3 =2 1 + 12 ×11 = 6 2 2x – 14𝑥 + 42 = 22 2x2 – 14𝑥 + 20 = 0 2x2 – 10𝑥 − 4𝑥 + 20 = 0 2x(x – 5) −4 (𝑥 − 5) = 0 x = 2, 5 From II alone we can determine the difference between blue & red balls in the bag. So, only statement II alone is sufficient to give answer of the question. 12 ×11 71 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S201. Ans.(e) Sol. √5776 - √1444 + √729 =43 + ? 76 – 38 +27 =43 + ? ?=65 -43 =22 S209. Ans.(d) Sol. 1 21 5 1 41 55 +? = 1 − 20 21 +? = 21 1440 1330 =√81 =9 S212. Ans.(c) Sol. 25×18+ 4200 40 − 525 105 = 740 − ? 450+105−5=740 −? ?= 740−550 =190 S213. Ans.(d) Sol. 3845+4380+2640 − 5965 = (?)2 (?)2=10865 – 5965 =4900 3 1 7 64 ?=√4900 =70 S214. Ans.(b) Sol. 400 ÷ 20 × 35 + 6666 ÷ 33+ ? = 1100 20× 35 + 202+? = 1100 ?=1100-(700+202) =1100- 902 =198 ? × 900 + 100 × 1050 = 100 × 3000 495 + 735 = 30 ×? 30 ×? = 1230 ? = 41 S208. Ans.(b) Sol. 73823 − 34156 + 4756 + 6758 − 9849 = 41499 − 160−? 41332 = 41339−? ?= 7 72 441 1 =√64 + 36 − 19 × 82 × 64 = 40 70 1 4 3840 S207. Ans.(a) Sol. 55% 𝑜𝑓 900 + 70% 𝑜𝑓 1050 = ? % 𝑜𝑓 3000 100 20 Sol.√ 60 + 40 − 70 ) ÷ (432 − 202 + 𝑜𝑓 21) × (82) =? 𝑜𝑓 × =? −62 S211. Ans.(b) 1 16 49 ?= 0 (16) ÷ (432 − 400 + 9) × (82) =?× 64 ?= 88 7 1 84 × × S206. Ans.(e) Sol. 5 × 1 S205. Ans.(b) Sol. 2450 +3760 -3830 =6000 - ? 2380 =6000 - ? ?=6000 -2380 = 3620 64 2036 Sol. 84 × 4 ÷ 212 +? = 147 × 21 − 21 S204. Ans.(c) Sol. 42.5×15 +37.5× 25= 1420 + ? 637.5+937.5 =1420 + ? ?= 1575 – 1420 = 155 𝑜𝑓25 3773 × S210. Ans.(c) S203. Ans.(a) Sol. 1484÷28 + 1462÷34 -12×7= ? ?=53+43 -84 = 12 4 1331 14 =? −36 ? = 50 S202. Ans.(a) Sol. 78 ×26÷6 +1262= 1311 + (?)2 2028÷6+1262 =1311 +(?)2 338+1262 =1311+(?)2 (?)2=1600 -1311 =289 ? =√289 =17 (5 5599 www.bankersadda.com | S215. Ans.(b) Sol. 28× 14.5+1680÷15+445=1000 -? 406+112+445=1000-? 963=1000-? ?=1000-963=37 www.sscadda.com | www.careerpower.in | Adda247 App S216. Ans.(c) Sol. 130 100 S223. Ans.(e) Sol. 13 × 900 √1296 + √2025 + √1521– √? = 1250 × 1200 + 50 × 30 = ? 100 130 × 12 + 25 × 30 = ? ? = 1560 + 750 ? = 2310 36 + 45 + 39 – √? = 117 √? = 120 – 117 ?=9 S217. Ans.(a) Sol. S224. Ans.(b) Sol. 56×240 350 + 14 + √? = (11)3 156 13 ? + (3)2 × 40 = 100 × 600 √? = 1331– 350 – 960 √? = 21 ? = 441 12 + 9 × 40 = ? × 6 372 ? = 6 = 62 S225. Ans.(c) Sol. 19×? 40 32 × 35 + √961 + 100 = 100 × 3305 S218. Ans.(a) Sol. 1 680 √81 × 36 + 17 = ? + (512)3 1120 + 31 + √2916 + 40 = ? + 8 ? = 54+ 40 – 8 = 86 19×? 100 × 140 + ?= 640 16 ? 100 × 1600 = 72 × 40 = 40 S220. Ans.(d) Sol. (17)2 + (22)2 + (8)2 + ? = 1750 − 820 + 2210 ? + 289 + 484 + 64 = 1750 – 820 + 2210 ? = 2303 S221. Ans.(a) Sol. 308 + 672 – 40 100 ×? + 80×355 100 2×? 2 S222. Ans.(b) Sol. 8 = –619 S227. Ans.(c) Sol. (575 + 7511 – 2769) ÷ (76 × 1 + 675 – 342) = √? = 5317 ÷ 409 = √? ⇒ ? = (13)2 = 169 S229. Ans.(d) Sol. = √(96) × 12 ÷ 18 + 26 – 9 = (65 – ? )% of 36 16 + 25 × 42 – 100 × 400 = (32)2 = 1024 + 64 – 1050 ? = 38 × 8 – 178 ? = 126 73 S226. Ans.(b) Sol. 1782 ÷ 54 + 456 – 2346 × 1 = ? × 3 ⇒ 33 + 456 – 2346 = ? × 3 ⇒ – 1857 = ? × 3 –1857 ⇒ ?= 3 1 ? = 1200 8 178+? 19 [(√3844 × 9) ÷ (27)3 ] × 23 =?2 + 337 ⇒ [(62 × 3) ÷ 3] × 23 =?2 + 337 ⇒ 1426 – 337 = ?² ⇒ ? = √1089 = 33 480×5 178+? = 1322 – 1151 S228. Ans.(a) Sol. = (28)2 980 + 284 – 784 = 5 ?= = 1322 ? = 900 16 × 140 + 16 × ? = 72 × 40 2240 + 16 × ? = 2880 ?= 100 100 171×100 S219. Ans.(e) Sol. 1600 19×? ⇒9= (65 –?) 100 × 36 ⇒ (65 – ? ) = ⇒ ? = 65 – 25 = 40 www.bankersadda.com | www.sscadda.com | 9×100 36 www.careerpower.in | Adda247 App S230. Ans.(a) Sol. 12 × √225 + 1212 – (1053 ÷ 9) = ? ⇒ 1392 – (117) = ? ⇒ ? = 1275 S238. Ans.(d) Sol. 35.99 × 4.98 − 1199.99 ÷ 7.99 =? 36 × 5 − 12 +? = =? S239. Ans.(b) Sol. ?2 + 60% 𝑜𝑓 239.99 = 55% 𝑜𝑓 320.02 + 3.98 60 S232. Ans.(d) Sol. √(123.09 + 465.05) ÷ 11.99+? = 240.02 ÷ 1.989 123+465 8 ? = 180 − 150 ? = 30 S231. Ans.(b) Sol. 42 28 × 350 − 100 × 400 =? 100 147 − 112 =? ? = 35 √ 1200 55 ?2 + 100 × 240 = 100 × 320 + 4 ?2 + 144 = 176 + 4 ?2 = 180 − 144 ?= 6 240 2 S240. Ans.(c) Sol. 524.90 + 125.05 =? × 9.99 525 + 125 =?× 10 √49+? = 120 ? = 113 S233. Ans.(e) Sol. (15.99)2 − 14.04 × 8.99+? = 154.999 162 − 14 × 9+? = 155 ? = 155 + 126 − 256 ? =25 650 ? = 10 ? = 65 S241. Ans.(e) Sol. √144.04 × 15% 𝑜𝑓 120.09 =? −54.99 × 3.03 S234. Ans.(a) Sol. 62 20 × 250 − × 105−? = 110 100 100 155 − 21 − 110 =? ? = 24 15 √144 × 100 × 120 =? −55 × 3 12 × 18 =? −165 ? = 216 + 165 ? = 381 S235. Ans.(c) Sol. 45% 𝑜𝑓 220.09 + 30% 𝑜𝑓 160.06 =?2 + 2.99 45 30 × 220 + 100 × 160 =?2 + 3 100 99 + 48 − 3 =?2 ? = 12 S242. Ans.(a) Sol. 13.03 × 7+? = 30.03% 𝑜𝑓 349.99 13 × 7+? = 30 100 × 350 91+? = 105 ? = 14 S236. Ans.(a) Sol. 1229.99 + 2120.09 − 3049.987 =? 1230 + 2120 − 3050 =? ? = 300 S237. Ans.(e) Sol. √√(99.99 + 104.99 × 5 =?÷ 8.989 √√100 + 105 × 5 = ? 9 √√625 = ? 9 ? = 45 74 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S243. Ans.(d) Sol. 32% 𝑜𝑓 600.02 − 19.99% 𝑜𝑓 400.04+? = 859.99 ÷ 2 32 100 20 860 × 600 − 100 × 400+? = 2 192 − 80+? = 430 ? = 318 S244. Ans.(b) Sol. 141 279.89 + − √? = 10.01 S249. Ans.(a) Sol. Wrong no. = 120 118 + 7 = 125 125 + 11 = 136 136 + 13 = 149 149 + 17 = 166 166 + 19 = 185 185 + 23 = 208 20.09 39.99 140 280 20 + 40 − √? = 10 √? = 7 + 7 − 10 ? = 16 S245. Ans.(c) Sol. 8.98 × 60.02 − 19.992 + 10.01% 𝑜𝑓 130.09 =? 10 9 × 60 − 202 + 100 × 130 =? 540 − 400 + 13 =? ? = 153 S250. Ans.(b) Sol. Wrong no. = 214 205 + 23 = 213 213 − 33 = 186 186 + 43 = 250 250 − 53 = 125 125 + 63 = 341 341 − 73 = −2 S246. Ans.(d) Sol. Wrong no. = 22 28 × 0.5 = 14 14 × 1 = 14 14 × 1.5 = 21 21 × 2 = 42 42 × 2.5 = 105 105 × 3 = 315 S251. Ans.(b) Sol. Wrong number = 810 Pattern of series – S247. Ans.(e) Sol. Wrong no. = 47 5 + (12 + 1) = 7 7 + (22 + 2) = 13 13 + (32 + 3) = 25 25 + (42 + 4) = 45 45 + (52 + 5) = 75 75 + (62 + 6) = 117 S252. Ans.(d) Sol. Wrong number = 350 Pattern of series – S248. Ans.(d) Sol. Wrong no. = 2400 288000 ÷ 12 = 24000 24000 ÷ 10 = 2400 2400 ÷ 8 = 300 300 ÷ 6 = 50 50 ÷ 4 = 12.5 12.5 ÷ 2 = 6.25 S253. Ans.(a) Sol. Wrong number = 646 Pattern of series – 75 814 820 832 + 12 +6 868 + 36 + 144 ×4 ×3 ×2 1732 1012 + 4320 + 720 ×5 6052 ×6 So, there should be 814 in place of 810. www.bankersadda.com 348 1024 832 -6 7 6 +484 (2 6 )² (2 2 )² 508 +196 -3 2 4 (1 4 )² (1 8 )² 604 704 640 -1 0 0 +36 (1 0 )² (6 )² So, there should be 348 in place of 350. 190 210 266 + 56 + 20 + 36 358 + 92 + 36 + 128 + 36 + 200 + 164 + 36 850 650 486 + 36 So, there should be 650 in place of 646. | www.sscadda.com | www.careerpower.in | Adda247 App S254. Ans.(e) S259. Ans.(a) Sol. Wrong number = 1 Pattern of series – Sol. Wrong number = 15 Pattern of series – 17 50 160 370 + 11 0 +33 +210 +160 1904 +696 +499 +339 +129 +100 +77 1208 709 +197 So, there should be 2 in place of 1. +29 +23 +31 +37 So, there should be 17 in place of 15. S260. Ans.(b) Sol. Wrong number = 1990 Pattern of series – S255. Ans.(b) Sol. Wrong number = 990 Pattern of series – 120 170 251 +81 +50 367 +4 +198 +43 +39 +35 +31 +155 + 11 6 722 522 +245 +47 So, there should be 1992 in place of 1990. +4 +4 +4 So, there should be 965 in place of 990. S261. Ans.(a) Sol. I: 2𝑥 2 + 𝑥 − 6 = 0 2𝑥 2 + 4𝑥 − 3𝑥 − 6 = 0 2𝑥(𝑥 + 2) − 3(𝑥 + 2) = 0 (2𝑥 − 3) (𝑥 + 2) = 0 𝑥 = 1.5, − 2 II: 𝑦 2 + 6𝑦 + 9 = 0 𝑦 2 + 3𝑦 + 3𝑦 + 9 = 0 𝑦(𝑦 + 3) + 3(𝑦 + 3) = 0 (𝑦 + 3)(𝑦 + 3) = 0 𝑦 = −3, −3 So, 𝑥 > 𝑦 S256. Ans.(b) Sol. Wrong number = 1090 Pattern of series – So, there should be 1080 in place of 1090. S257. Ans.(e) Sol. Wrong number = 110 Pattern of series – S262. Ans.(d) Sol. I: 𝑥 2 − 4𝑥 + 4 = 0 𝑥 2 − 2𝑥 − 2𝑥 + 4 = 0 𝑥(𝑥 − 2) − 2(𝑥 − 2) = 0 (𝑥 − 2)(𝑥 − 2) = 0 𝑥 = 2, 2 II: 𝑦 2 − 10𝑦 + 16 = 0 𝑦 2 − 8𝑦 − 2𝑦 + 16 = 0 𝑦(𝑦 − 8) − 2(𝑦 − 8) = 0 (𝑦 − 8)(𝑦 − 2) = 0 𝑦 = 8, 2 So, 𝑥 ≤ 𝑦 So, there should be 113 in place of 110. S258. Ans.(d) Sol. Wrong number = 255 Pattern of series – So, there should be 256 in place of 255. 76 965 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S263. Ans.(e) Sol. I: 2𝑥 2 + 7𝑥 + 6 = 0 2𝑥 2 + 3𝑥 + 4𝑥 + 6 = 0 𝑥(2𝑥 + 3) + 2(2𝑥 + 3) = 0 (2𝑥 + 3)(𝑥 + 2) = 0 S266. Ans.(a) Sol. I: 4𝑥 2 − 20𝑥 + 25 = 0 4𝑥 2 − 10𝑥 − 10𝑥 + 25 = 0 2𝑥(2𝑥 − 5) − 5(2𝑥 − 5) = 0 (2𝑥 − 5)(2𝑥 − 5) = 0 5 5 3 𝑥 = − 2 , −2 𝑥 = 2,2 II: 3𝑦 2 + 11𝑦 + 10 = 0 3𝑦 2 + 6𝑦 + 5𝑦 + 10 = 0 3𝑦(𝑦 + 2) + 5(𝑦 + 2) = 0 (3𝑦 + 5)(𝑦 + 2) = 0 II: 5𝑦 2 − 6𝑦 − 8 = 0 5𝑦 2 − 10𝑦 + 4𝑦 − 8 = 0 5𝑦(𝑦 − 2) + 4(𝑦 − 2) = 0 (5𝑦 + 4)(𝑦 − 2) = 0 4 5 𝑦 = 2, − 5 𝑦 = − 3 , −2 So, no relation can be established. So, 𝑥 > 𝑦 S264. Ans.(d) Sol. I: 𝑥 2 − 2𝑥 − 24 = 0 𝑥 2 − 6𝑥 + 4𝑥 − 24 = 0 𝑥(𝑥 − 6) + 4(𝑥 − 6) = 0 (𝑥 − 6)(𝑥 + 4) = 0 𝑥 = −4, 6 II: 𝑦 2 − 12𝑦 + 36 = 0 𝑦 2 − 6𝑦 − 6𝑦 + 36 = 0 𝑦(𝑦 − 6) − 6(𝑦 − 6) = 0 (𝑦 − 6)(𝑦 − 6) = 0 𝑦 = 6, 6 So, 𝑥 ≤ 𝑦 S267. Ans.(b) Sol. I: 𝑥 2 − 2𝑥 − 15 = 0 𝑥 2 − 5𝑥 + 3𝑥 − 15 = 0 𝑥(𝑥 − 5) + 3(𝑥 − 5) = 0 (𝑥 + 3) (𝑥 − 5) = 0 𝑥 = −3, 5 II: 𝑦 2 − 15𝑦 + 56 = 0 𝑦 2 − 8𝑦 − 7𝑦 + 56 = 0 𝑦(𝑦 − 8) − 7(𝑦 − 8) = 0 (𝑦 − 7)(𝑦 − 8) = 0 𝑦 = 7, 8 So, 𝑥 < 𝑦 S265. Ans.(a) Sol. I: 4𝑥 2 + 11𝑥 + 6 = 0 4𝑥 2 + 8𝑥 + 3𝑥 + 6 = 0 4𝑥(𝑥 + 2) + 3(𝑥 + 2) = 0 (4𝑥 + 3)(𝑥 + 2) = 0 S268. Ans.(e) Sol. I: 10𝑥 2 + 19𝑥 + 7 = 0 10𝑥 2 + 14𝑥 + 5𝑥 + 7 = 0 2𝑥(5𝑥 + 7) + 1(5𝑥 + 7) = 0 (2𝑥 + 1)(5𝑥 + 7) = 0 1 7 2 5 𝑥 = − , −2 𝑥 = − ,− II: 𝑦 2 + 10𝑦 + 25 = 0 𝑦 2 + 5𝑦 + 5𝑦 + 25 = 0 𝑦(𝑦 + 5) + 5(𝑦 + 5) = 0 (𝑦 + 5)(𝑦 + 5) = 0 𝑦 = −5, −5 So, 𝑥 > 𝑦 II: 5𝑦 2 + 16𝑦 + 12 = 0 5𝑦 2 + 6𝑦 + 10𝑦 + 12 = 0 𝑦(5𝑦 + 6) + 2(5𝑦 + 6) = 0 (𝑦 + 2)(5𝑦 + 6) = 0 3 4 77 www.bankersadda.com 6 𝑦 = −2, − 5 So, no relation can be established. | www.sscadda.com | www.careerpower.in | Adda247 App S269. Ans.(a) Sol. I: 𝑥 2 − 20𝑥 + 75 = 0 𝑥 2 − 15𝑥 − 5𝑥 + 75 = 0 𝑥(𝑥 − 15) − 5(𝑥 − 15) = 0 (𝑥 − 5)(𝑥 − 15) = 0 𝑥 = 5, 15 II: 𝑦 2 + 19𝑦 + 84 = 0 𝑦 2 + 12𝑦 + 7𝑦 + 84 = 0 𝑦(𝑦 + 12) + 7(𝑦 + 12) = 0 (𝑦 + 12)(𝑦 + 7) = 0 𝑦 = −12, −7 So, 𝑥 > 𝑦 S272. Ans.(c) Sol. I: 𝑥 2 − 18𝑥 + 56 = 0 𝑥 2 − 14𝑥 − 4𝑥 + 56 = 0 𝑥(𝑥 − 14) − 4(𝑥 − 14) = 0 (𝑥 − 4)(𝑥 − 14) = 0 𝑥 = 4, 14 II: 𝑦 2 + 4𝑦 − 32 = 0 𝑦 2 + 8𝑦 − 4𝑦 − 32 = 0 𝑦(𝑦 + 8) − 4(𝑦 − 8) = 0 (𝑦 − 4)(𝑦 + 8) = 0 𝑦 = −8, 4 So, 𝑥 ≥ 𝑦 S273. Ans.(d) Sol. I: 𝑥 2 + 14𝑥 − 72 = 0 𝑥 2 + 18𝑥 − 4𝑥 − 72 = 0 𝑥(𝑥 + 18) − 4(𝑥 + 18) = 0 (𝑥 + 18)(𝑥 − 4) = 0 𝑥 = −18, 4 II: 𝑦 2 − 13𝑦 + 36 = 0 𝑦 2 − 9𝑦 − 4𝑦 + 36 = 0 𝑦(𝑦 − 9) − 4(𝑦 − 9) = 0 (𝑦 − 4)(𝑦 − 9) = 0 𝑦 = 4, 9 So, 𝑥 ≤ 𝑦 S270. Ans.(e) Sol. I: 𝑥 2 − 9𝑥 − 22 = 0 𝑥 2 − 11𝑥 + 2𝑥 − 22 = 0 𝑥(𝑥 − 11) + 2(2𝑥 − 11) = 0 (𝑥 + 2)(𝑥 − 11) = 0 𝑥 = −2, 11 II: 𝑦 2 − 17𝑦 + 66 = 0 𝑦 2 − 11𝑦 − 6𝑦 + 66 = 0 𝑦(𝑦 − 11) − 6(𝑦 − 11) = 0 (𝑦 − 11)(𝑦 − 6) = 0 𝑦 = 6, 11 So, no relation can be established. S274. Ans.(d) Sol. I: 𝑥 2 − 92 = 122 𝑥 2 = 144 + 81 𝑥 2 = 225 𝑥 = 15, −15 II: 𝑦 3 = 3375 𝑦 = 15 So, 𝑥 ≤ 𝑦 S271. Ans.(b) Sol. I: 4𝑥 2 + 19𝑥 + 15 = 0 4𝑥 2 + 15𝑥 + 4𝑥 + 15 = 0 𝑥(4𝑥 + 15) + 1(4𝑥 + 15) = 0 (4𝑥 + 15)(𝑥 + 1) = 0 𝑥 = −1, −15 II: 8𝑦 2 + 10𝑦 + 3 = 0 8𝑦 2 + 6𝑦 + 4𝑦 + 3 = 0 2𝑦(4𝑦 + 3) + 1(4𝑦 + 3) = 0 (4𝑦 + 3)(2𝑦 + 1) = 0 3 S275. Ans.(b) Sol. I: 5 𝑥2 28 5 3 1 So, 𝑥 < 𝑦 www.bankersadda.com 𝑥2 7 28 𝑥 2− 2 = 7 𝑥=4 II: 11𝑦 + (7 × 6) = 97 11𝑦 + 42 = 97 11𝑦 = 55 𝑦 =5 So, 𝑥 < 𝑦 𝑦 = −4,−2 78 3 = | www.sscadda.com | www.careerpower.in | Adda247 App S276. Ans.(a) Sol. I. 𝑥 2 − 22𝑥 + 72 = 0 x2 -18x -4x +72=0 x(x-18)-4(x-18)=0 (x-18)(x-4)=0 x = 18, 4 II. 𝑦 2 + 11𝑦 + 30 = 0 y2+5y +6y +30=0 y(y+5) +6(y+5)=0 (y+5)(y+6)=0 y= -5,-6 So, x>y S280. Ans.(c) Sol. I. 𝑥 2 + 4𝑥 − 12 = 0 x2+6x-2x-12=0 x(x+6)-2(x+6)=0 (x+6) (x-2)=0 x=-6, 2 II. 𝑦 2 − 9𝑦 + 20 = 0 y2-5y-4y+20=0 y(y-5)-4(y-5)=0 (y-5) (y-4)=0 y=5, 4 𝑠𝑜, 𝑦 > 𝑥 S277. Ans.(e) Sol. I. 𝑥 2 − 23𝑥 + 120 = 0 x2 -15x -8x +120=0 x(x-15)- 8(x-15)=0 (x-15)(x-8)=0 x = 15, 8 II. 𝑦 2 − 17𝑦 + 70 = 0 y2-10y -7y +70=0 y(y-10) -7(y-10)=0 (y-7)(y-10)=0 y= 10, 7 So, no relation can be established S281. Ans.(e) Sol. I. 𝑥 2 − 25𝑥 + 100 = 0 x2 -20x -5x +100=0 x(x-20)-5(x-20)=0 (x-20) (x-5)=0 x =20,5 II. 𝑦 2 − 27𝑦 + 110 = 0 y2-22y -5y +110=0 y(y-22) -5(y-22)=0 (y-22)(y-5)=0 y=22,5 So, no relation can be established between x and y. S278. Ans.(b) Sol. I. 𝑥 2 − 15𝑥 + 54 = 0 x2-9x-6x+54=0 x(x-9)-6(x-9)=0 (x-9)(x-6)=0 x= 9, 6 II. 𝑦 2 + 10𝑦 − 96 = 0 y2+16y-6y-96 =0 y(y+16)-6(y+16)=0 (y+16)(y-6)=0 y = 6,−16 So, 𝑥 ≥ 𝑦 S282. Ans.(d) Sol. I. 𝑥 2 = 289 x =√289 x =17, −17 II. 𝑦 = √289 y =17 So, x ≤ y S279. Ans.(b) Sol. I.x3 +440=2168 x3=2168-440 x3=1728 x =12 II. 𝑦 2 − 23 =121 y2=121+23 y2=144 y =12, −12 So, x ≥ y 79 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S287. Ans.(b) Sol. I. 𝑥 2 − 14𝑥 + 45 = 0 𝑥 2 − 9𝑥 − 5𝑥 + 45 = 0 𝑥(𝑥 − 9) − 5(𝑥 − 9) = 0 (𝑥 − 9)(𝑥 − 5) = 0 𝑥 = 9, 5 II. 𝑦 2 + 2𝑦 − 35 = 0 𝑦 2 + 7𝑦 − 5𝑦 − 35 = 0 𝑦(𝑦 + 7) − 5(𝑦 + 7) = 0 (𝑦 − 5)(𝑦 + 7) = 0 𝑦 = 5, −7 So, 𝑥 ≥ 𝑦 S283. Ans.(d) Sol. I. 𝑥 2 + 12𝑥 + 32 = 0 x2+8x+4x+32=0 x(x+8)+4(x+8)=0 (x+8)(x+4)=0 x= −8, −4 II. 𝑦 2 + 7𝑦 + 12 = 0 y2+ 3y+4y+12 =0 y(y+3)+4(y+3)=0 (y+4)(y+3)=0 y = −4, −3 So, 𝑦 ≥ 𝑥 S284. Ans.(a) Sol. I. 3𝑥² + 16𝑥 + 20 = 0 3x2 +6x +10x +20 =0 3x(x+2) +10(x+2)=0 (3x+10)(x+2)=0 10 x= −2, − 3 II. 𝑦² + 14𝑦 + 48 =0 y2 +8y +6y +48 =0 y(y +8)+6(y+8)=0 (y+6)(y+8)=0 y =−6, −8 So, x > y 𝐒𝟐85. Ans.(e) Sol. I. 𝑥 2 + 𝑥 − 72 = 0 x2+ 9x-8x-72=0 x(x+9)-8(x+9)=0 (x+9)(x-8)=0 x = 8,−9 II. 𝑦 2 + 13𝑦 + 42=0 y2 +6y+7y+42 =0 y(y+6)+7(y+6)=0 (y+6)(y+7)=0 y = -6,-7 So, no relation can be established between x and y. S286. Ans.(c) Sol. I. 𝑥 2 + 5𝑥 + 6 = 0 𝑥 2 + 3𝑥 + 2𝑥 + 6 = 0 𝑥(𝑥 + 3) + 2(𝑥 + 3) = 0 (𝑥 + 3)(𝑥 + 2) = 0 𝑥 = −2, −3 II. 𝑦 2 − 9𝑦 + 14 = 0 𝑦 2 − 7𝑦 − 2𝑦 + 14 = 0 𝑦(𝑦 − 7) − 2(𝑦 − 7) = 0 (𝑦 − 2)(𝑦 − 7) = 0 𝑦 = 2, 7 So, y>x 80 www.bankersadda.com S288. Ans.(e) Sol. I. 𝑥 2 + 11𝑥 + 18 = 0 𝑥 2 + 9𝑥 + 2𝑥 + 18 = 0 𝑥(𝑥 + 9) + 2(𝑥 + 9) = 0 (𝑥 + 9)(𝑥 + 2) = 0 𝑥 = −2, −9 II. 𝑦 2 + 6𝑦 + 8 = 0 𝑦 2 + 4𝑦 + 2𝑦 + 8 = 0 𝑦(𝑦 + 4) + 2(𝑦 + 4) = 0 (𝑦 + 4)(𝑦 + 2) = 0 𝑦 = −4, −2 So, no relation can be established S289. Ans.(c) Sol. I. 𝑥² + 5𝑥 + 6 = 0 𝑥 2 + 3𝑥 + 2𝑥 + 6 = 0 𝑥(𝑥 + 3) + 2(𝑥 + 3) = 0 (𝑥 + 3)(𝑥 + 2) = 0 𝑥 = −3, −2 II. 𝑦 2 – 15𝑦 = 16 𝑦 2 − 15𝑦 − 16 = 0 𝑦 2 − 16𝑦 + 𝑦 − 16 = 0 𝑦(𝑦 − 16) + 1(𝑦 − 16) = 0 (𝑦 + 1)(𝑦 − 16) = 0 𝑦 = −1, 16 Clearly, 𝑥 < 𝑦 S290. Ans.(a) Sol. Multiplying II by 3 and subtracting II from I, we get, 𝑦 = −1 𝑎𝑛𝑑 𝑥 = 3 So, 𝑥 > 𝑦 S291. Ans.(b) Sol. Required percentage = (22+23)−10 22+23 × 100 ≈ 78% | www.sscadda.com | www.careerpower.in | Adda247 App S292. Ans.(c) S299. Ans.(b) Sol. Population which is literate but ungraduated from 1 Sol. Required average = 3 × (12 + 15 + 18)% × 22500 city A = 10000 × 15 = 100 × 22500 60 100 = 6000 6000 Required percentage = 2400 × 100 = 250% = 3375 S293. Ans.(d) S300. Ans.(c) 23−12 Sol. Required no. of passenger = 22500 × 100 = 2475 Sol. Graduate male from city C = S294. Ans.(d) Literate but ungraduated from city B = 6000 × 100 = 3600 25600 16 × 9 = 14400 60 10 7 23 5 Sol. Required ratio = 22500 × 100 × 15 : 22500 × 100 × 23 = 14: 15 Required difference = 14400 − 3600 = 10800 S301. Ans.(b) 2200 Sol. Passenger travelling to Rewari = 22500 × 18 100 = 4050 15 Passenger travelling to Panipat = 22500 × 100 = 3375 4 1200 Selling price of per kg sugar = 40 = 𝑅𝑠. 30 Required difference = 30 − 25 = 𝑅𝑠. 5 less Required difference = 3375 × 75 × − 4050 × 75 S302. Ans.(e) = 75 × (4500 − 4050) = 75 × 450 = 33750 Rs. Sol (96-100): For city C Sol. selling price of one kg wheat = 3 Required average selling price = 2 20×3+10×2 80 = 5 3+2 S303. Ans.(b) 900 Sol. Required percentage = 2200−600 × 100 = 56.25% 1 Illiterate population = 96000 × = 32000 3 40 = 𝑅𝑠. 20 = Rs. 16 per kg Literate population of city C = 96000 × 3 = 64000 100 45 Selling price of one kg salt = 60 = 𝑅𝑠. 10 6000 Graduate population = 64000 × 900 600 Total population of city C = 6.25 × 100 = 96000 = 25600 S304. Ans.(b) For city B Total population = 16000 Literate population = 6000 Illiterate population = 16000 − 6000 = 10000 Sol. selling price of a kg pulse = 3750 50 = 𝑅𝑠. 75 Profit earned on selling of one kg pulse = 75 − 60 = 𝑅𝑠. 15 Total profit = 15 × 40 = 𝑅𝑠. 600 40 Graduate population = 6000 × 100 = 2400 For city A Total population = 22000 S305. Ans.(e) 1 Sol. Required average quantity = 3 × (55 + 50 + 45) 5 Literate population = 22000 × 11 = 10000 150 = 3 Illiterate population = 22000 − 10000 = 12000 40 = 50 kg Graduate population = 10000 × 100 = 4000 S306. Ans.(a) Sol. No. of male student playing Hockey of college L S296. Ans.(c) 6000 Sol. Required percentage = 12000 × 100 = 50% 8 = 450 × 9 = 400 S297. Ans.(d) Sol. Required ratio = 25600: 16000 = 8:5 Average no. of student playing Hockey of college M & O = S298. Ans.(a) Sol. Required difference = 32000 − 2400 = 29600 81 100 Sol. Cost price of per kg rice = 55 × 160 = 𝑅𝑠. 25 S295. Ans.(a) www.bankersadda.com | 400+500 2 = 450 400 8 Required percentage = 450 × 100 = 88 9 % www.sscadda.com | www.careerpower.in | Adda247 App S307. Ans.(c) Sol. Student who left playing Cricket of college N S311. Ans.(b) Sol. USA has lowest mortality rate, which is 3.14% 1 = 350 × 7 = 50 S312. Ans.(d) Total student playing Football of college N = 450 + 50 = 500 Required ratio = 500+300 500+300 Sol. Required % = =1∶ 1 (250+450+500) 3 = 400 Average no. of student playing Football of college K, L and M = 400+350+300 3 = 350 650 × 100 = 7 = 15 : 7 S314. Ans.(a) Sol. Required % = 4000+11000+17500+15000 80000 × 100 = 59.375% 5 9 13 4 100000 Mortality rate in china is 5%. New number of total deaths = 100000× 5 100 = 5000 S316. Ans.(e) 11 Sol. Price of a one kg sugar = 84 × 21 = 𝑅𝑠 44 10 Price of one kg of salt = 840 × 21 = 𝑅𝑠40 % S310. Ans.(d) Sol. Total student in college K in 2014 = 400 + 500 + 250 = 1150 Total student in college K in 2015 120 = 1150 × 100 = 1380 Student playing Football of college K in 2015 Required difference = (20 × 44 – 15 × 40) = 880 – 600 = 𝑅𝑠. 280 S317. Ans.(a) Sol. Price of one kg of tea = Price of one kg of rice = 5 = 1380 × 10 Required % = = 690 400+690 Required average = 2 = 80000 ×100 ×100 Sol. New total confirmed cases in china = 80000 × = S309. Ans.(e) Sol. Total no. of student playing Cricket of college L and M together = 400 + 300 = 700 Total no. of student playing Hockey of college K and M together = 250 + 400 = 650 700–650 15000 Sol. Required ratio = 140000 4000 S315. Ans.(e) Required difference = 400 – 350 = 50 Required percentage = × 100 = 1900% 17500 S313. Ans.(c) S308. Ans.(b) Sol. Average no. of student playing Hockey of college K, L and O = 350000−17500 50−50 50 900 1500 30 18 = 𝑅𝑠50 = 𝑅𝑠50 × 100 = 0% S318. Ans.(d) 1090 2 63×12 18 = 545 Sol. Required ratio = 42×25 = 25 Solution (111-115): ATQ, 4000 Mortality rate for China = 80000 × 100 = 5% S319. Ans.(b) 20+15 1 Required%= 30+12 × 100 = 83 3 % 11000 Mortality rate for USA = 350000 × 100 = 3.14% 17500 Mortality rate for Italy = 130000 × 100 = 13.46% Mortality rate for Spain = 82 15000 140000 × 100 = 10.71% www.bankersadda.com | S320. Ans.(b) Sol. Required sum = (56 × 15) + (32 × 30) + (40 × 25) = 2800 Rs. www.sscadda.com | www.careerpower.in | Adda247 App S321. Ans.(c) Sol. Let length of train A = l metres. And let speed of train A = S m/s. ATQ, Speed of train B = S325. Ans.(b) Sol. let speed of boat in still water and speed of Stream be P and Q kmph respectively. ATQ, 450+150 Speed of train A, S = 𝑙+230 450+𝑙 450+𝑙 160 = 25 − S326. Ans.(c) Sol. Given distance between P and Q is 900 km. 𝑙+230 l=350 metres. = 8 𝑘𝑚𝑝ℎ (Upstream Speed) …(ii) 160 On solving (i) & (ii): 𝑙+450 5 P+Q = 16 𝑘𝑚𝑝ℎ (Downstream Speed) ATQ, Downstream Speed, X-4 = P+Q So, X = 16+4 = 20. …(i) 29 Now, 25 − 𝑆 = 160 𝑆 = 25 − 40 P-Q = 24 = 25 m/s 29 So, speed of train A = 350+230 = 20 m/s. Required time = speed of car B = 29 20 S322. Ans.(e) Sol. Let the speed of train A and train B be 17X m/s and 13X m/s respectively. And let the length of train B = Y meter 950+𝑌 ATQ, 17𝑋−13𝑋 = 16 Y = 64X -950, So, length can’t be determined with given data. 900 (𝑋+4) 9 900 5 ×2= 𝑋 ×2 9X = 5 (X+4) 4X = 20 So, speed of car B = 900 (5+4) =100 kmph. 9 Required distance= 100 × = 450 km 2 S327. Ans.(b) Sol. Now, let speed of the boat in still water and the speed of the stream be a km/hr. & b km/hr. respectively. So, upstream speed of boat = (𝑎 − 𝑏) km/hr. ATQ, 𝑎 − 𝑏 = 15 5 3𝐿 = 144 × 18 × 30 L = 400 m Length of train = 800 m ∴ Length of other train = 2 × 800 = 1600 m 5 ATQ, Car B started from P at 6:00am and car A started from P at 8:00 am They both met at 10:30 am i.e. X = 5 hours S323. Ans.(d) Sol. Let length of train = 2L m Length of tunnel = L m ATQ, 60 60% of speed = 144 × 18 × 100 = 24 m/sec. 120 ∴ (1600 + 800) = 24 × time ∴ time = 100 sec. Required time= (𝑎−𝑏) S324. Ans.(b) Sol. Let us assume the original speed of Deepak be 4x km/hr and original time taken by Deepak be T hr. ATQ, decreased speed of Deepak = 3x km/hr, =8 hr. = 24 And increased time of Deepak = (T+60) = (𝑇 + 0.40) hours So, 4𝑥 × 𝑇 = 3𝑥 × (𝑇 + 0.4) T = 1.2hour = 72 minutes 83 km/h. Speed of car A = 𝑋 km/h. 350+50 = 20 sec. 900 (𝑋+4) 900 www.bankersadda.com 120 15 S328. Ans.(c) Sol. Let the distance between Amit’s home and his office is D km. 𝐷 𝐷 2𝐷 ATQ, 30 + 𝑋 = 33 X = 36.67 km/hr | www.sscadda.com | www.careerpower.in | Adda247 App S329. Ans.(a) Sol. Time taken by X = 8 hr. Time taken by Y = 7 hr. S332. Ans.(b) Sol. Let speed of stream = r km/h A/q, (8 − 𝑟) × 5 = (8 + 𝑟) × 3 ⇒ 40 – 5r = 24 + 3r 16 ⇒ 𝑟 = 8 = 2 𝑘𝑚/ℎ ∴ time taken to cross each other 56 S333. Ans.(b) Sol. Let total distance = d 11 = 15 = 3 15 hr. = 3 hr 44 min. ∴ Required time to cross = 11 : 44 am ∴ Average speed = 𝑑 S330. Ans.(b) Sol. Let initial speed of the car = s kmph. And initial time taken by the car to cover the distance = t hours. So, Total Distance = 𝑠 × 𝑡 km. ATQ, (𝑠 − 9)(𝑡 + 2) = (𝑠 + 5) (𝑡 − 48 60 ) s-5t = 5 …..(i) and, st = (s-9) (t+2) 2s-9t = 18 …….(ii) From eq(i) & eq(ii) t=8 hours and s= 45 kmph so, required distance = 45 × 8 = 360 𝑘𝑚. 1100 700 S334. Ans.(a) Sol. Let the total distance = x km x x 90 + = 12– 4 12 + 4 60 x x + = 1.5 8 16 3x = 1.5 × 16 x = 8km 2 ⇒ x = 7t Let t = 7y = time taken on train x = 2y = time taken on car t−x = 5y = time taken on cycle. Required Ratio → 60 × 2y : 4 × 5y : 20 × 7y 6 : 1 : 7 × 7𝑥 = 11x km/hr. Hence, speed of boat in still water = 𝑑 + 24 48 S335. Ans.(d) Sol. Let one side time taken = t hour Time taken by car = x hour ATQ, 60x + 4(t – x) = 20 × t S331. Ans.(b) Sol. Let upstream speed of a boat be 7x km/hr. So, downstream speed of a boat = = 16 km/h 𝑑 7𝑥+11𝑥 2 = 9x km/hr. And, speed of stream = 11𝑥 − 9𝑥 = 2x km/hr. ATQ, 2𝑥 = 8 𝑥=4 176 70 Required time = 11𝑥 + 7𝑥 16 10 = 𝑥 +𝑥 26 = 𝑥 = 6.5 hours 84 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S336. Ans.(c) Sol. In both conical shape volume will be same. Let base radius of cone is R cm So, height of the cone = 2R cm. ATQ, 4 3 S341. Ans.(d) Sol. Total number of cases when two dices are rolled simultaneously=36 total cases of getting same number on both the dices=(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 6 5 required probability=1 − 36 = 6 1 𝜋(16)3 = 2 × 𝜋𝑅2 × 2𝑅 3 𝑅3 = 163 R =16 Required height = 2R =32 cm. S342. Ans.(d) Sol. 4 Volume of sphere = 𝜋𝑅3 (R → Radius) 3 S337. Ans.(d) Sol. According to question the first place of the three-letter word will be fix & will be filled by S only. So, rest two letter will be selected from the rest 6 letter of word STRANGE. So, Number of possible ways = 6 × 5 = 30 S338. Ans.(b) Sol. total outcomes = 62 = 36 Favorable outcomes = when sum is 2, 3, 4, 8, 9, 10 (1,1) (1,2) (1,3) (2,1) (2,2) (2,6) (3,1) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (6,2) (6,3) (6,4) Required probability = 18 36 = 2 1 2 × 𝑎 × 𝑎 = 2048 m2 a2 = 4096. a=64 m. so, its hypotenuse = 64√2 m. 7.5×7.5×7.5 S345. Ans.(a) Sol. As we know there exist 2 black queens and 2 kings in a set of 52 playing cards. 2C 1 = 8√2 𝑚. Total surface area of the sphere = 4𝜋 × 8√2 × 8√2 =512𝜋 m2 S340. Ans.(d) Sol. Let number of employees in ‘Adda 247’ initially = n ATQ – = (5 + 1) 5n + 26 = 6n + 6 n = 20 New number of employees in ‘Adda 247’ = 20 + 1= 21 www.bankersadda.com S344. Ans.(b) Sol. ATQ, vowels have to come together so A and I together will be treated as a single letter. And, A and I can change their respective places in 2! Ways. So, Number of ways = (8-1)! × 2!= 7!× 2! = 10080 ways 2C 1 85 S343. Ans.(c) 45×45×45 Sol. Number of cubes = = 216 1 So, Required Probability = 52C1 + 52C1 = 13 Now, radius of the Sphere = 8 × 64√2 (𝑛+1) Radius of cylinder=r=6cm Height of cylinder=h=12cm Volume of cylinder = πr²h = 432π cm³ 1 S339. Ans.(a) Sol. Let each of base and height of the isosceles right-angle triangle is a meter so its hypotenuse will be a√2 m. Area of isosceles right-angle triangle = 128 × 16 (5𝑛+26) Volume of cylinder = πr²h (r → radius of cylinder, h → height of cylinder) R = r (given) ATQ, 4 𝜋𝑅3 =288𝜋 ⇒ R3 =216 ⇒ R=6cm=r 3 | 1 S346. Ans.(d) Sol. Let the radius of cylinder and hemisphere be r cm. So, height of cylinder = 2r cm. Surface area of cylinder = 2πrh = 4πr2 Total Surface Area of Hemi-Sphere = 3πr2 Required result = 1 4π𝑟2 −3π𝑟2 3π𝑟2 × 100 =33 3 % S347. Ans.(e) Sol. Possible cases of balls will be 2 red or 2 Orange or 2 Green. 4C 3C 2C 6 3 1 5 Required probability = 9C2 + 9C2 + 9C2 = 36 + 36 + 36 = 18 www.sscadda.com 2 | 2 www.careerpower.in 2 | Adda247 App S348. Ans.(a) Sol. In the word BLASTING, there are two vowels (A, I) and six consonants (B, L, S, T, N, G). So, required probability = 7!×2! 8! 2 1 =8=4 S349. Ans.(c) Sol. radius = r cm Height =3r cm ATQ 2πr(r+h)=1232 ⟹ 2× 22 7 S356.Ans.(c) Sol. Wrong number = 10 Pattern of series – 8 × 0.5 = 4 4×1=4 4 × 1.5 = 𝟔 𝟔 × 2 = 12 12 × 2.5 = 30 30 × 3 = 90 × 𝑟 × 4𝑟 = 1232 ⟹ r = 7cm Height = h=21cm 22 Volume of cylinder = 7 × 7 × 7 × 21 = 3234 𝑐𝑚3 S350. Ans.(e) Sol. Let us suppose number of green balls in the box = x ATQ, 6C 1 (6+5+x)C 6 𝑥+11 1 1 =3 = 1 3 x+11 =18 ∴x=7 S351.Ans.(a) Sol. Number of complaints received Tuesday= 100 + 80 + 70 + 110 =360 Number of complaints received on Wednesday= 50 + 60 + 120 + 90 = 320 Required difference= 360 − 320 = 40 S352.Ans.(b) Sol. Required %= (70+110)−(50+60) (50+60) 70 = 110 × 100 = 63.63 S354.Ans.(c) Sol. (70+120) 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 % = (100+80+70+110) × 100 190 = 360 × 100 = 52.77 ≈ 53 % www.bankersadda.com S357. Ans.(c) Sol. Wrong number = 11 Pattern of series – 𝟏𝟐 + 22 = 16 16 + 32 = 25 25 + 42 = 41 41 + 52 = 66 66 + 62 = 102 102 + 72 = 151 S358. Ans.(d) Sol. Wrong number = 25 Pattern of series – 21 + 23 = 𝟐𝟗 𝟐𝟗 − 32 = 20 20 + 23 = 28 28 − 32 = 19 19 + 23 = 27 27 − 32 = 18 S359.Ans.(a) Sol. Wrong number = 104 Pattern of series – 20 + 8 = 28 28 + 12 = 40 40 + 16 = 56 56 + 20 = 76 76 + 24 = 𝟏𝟎𝟎 𝟏𝟎𝟎 + 28 = 128 S353.Ans.(c) Sol. 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑎𝑡𝑖𝑜 = (80 + 60): (50 + 90) = 1: 1 86 S355. Ans.(d) Sol. Required ratio=(100 + 80 + 110): (50 + 60 + 120 + 90) = 290: 320 = 29: 32 | www.sscadda.com | www.careerpower.in | Adda247 App S360.Ans.(e) Sol. Wrong number = 20 Pattern of series – 1×1+1= 2 2×2+2= 6 6 × 3 + 3 = 𝟐𝟏 𝟐𝟏 × 4 + 4 = 88 88 × 5 + 5 = 445 445 × 6 + 6 = 2676 S368. Ans.(a) Sol. 4? + 4? = 256 ?=4 S369. Ans.(e) Sol. 1512 ? 1512 ? ? 100 ? 100 ?= 288 ×100 640 (80x−70x) 80x × 100 S373. Ans.(b) Sol. Let total production in any of these years be 100x 62.5 ? = 400 S366. Ans.(b) Sol. 30+60+30 % of 100x 3 Required ratio = (30+30)% = 2: 3 of 100x × 450 + ?2 = 256 – 4 S374. Ans.(d) Sol. Let total production in any of these years be 100x ATQ, 10% of 100x = 12000 x = 1200 10% of 1,20,000 + 30% of 1,20,000 Required average = 2 = 24000 ?2 = 252 – 108 ? = 12 S367. Ans.(d) Sol. × 750 + 110) = 88 100 × 2500 ? × 440 = 2200 ?=5 87 × 640 = 400 – 112 S372. Ans.(e) Sol. Required difference = 60% of 1,50,000 – (20+30) % of 1,50,000 = 15000 250×100 44 40 × 640 + 100 × 280 = 400 = 12.5 % ×? −25 = 225 100 × 400 = 280 – 244 ∴ Required percent = S365.Ans.(d) Sol. ?×( 70 100 S371. Ans.(d) Sol. Let total production in any of these years be 100x S364.Ans.(e) Sol. 3167 − 2881 − 121 =? −41 ? = 206 24 × 488 = ? = 45 S363.Ans.(c) Sol. 75 − 63 + 25 =? ? = 37 100 50 100 S370.Ans(d) Sol. 43−16 = √?− 12 ? = 1521 ?= + ? = 42 S362.Ans.(e) Sol. 100 × 980 = 1040 4? + 784 = 1040 S361.Ans.(a) Sol. 63 + 18 =?2 ?= 9 62.5 80 100 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App S375. Ans.(b) Sol. Let total production in any of these years be 100x 20 ∴ × 100x = 18000 100 x=900 Total production in 2019 = = 1,08,000. 120 100 S384. Ans.(d) Sol. Sum of selling price of all 3 type of refrigerators in year 2016 = (16000+22000+26000) = 64000 Sum of selling price of all 3 type of refrigerators in year 2017 = (14000+25000+32000) = 71000 Sum of selling price of all 3 type of refrigerators in year 2018 = (15000+19000+29000) = 63000 Sum of selling price of all 3 type of refrigerators in year 2019 = (17000+22000+28000) = 67000 So, in Year 2018 it is lowest. × 100 × 900 S376. Ans.(b) Sol. –26, +52, –78, +104, –130 So, 640 – 130 = 510 S385. Ans.(b) Sol. 15000 Required % = 17000 × 100 = 88.23% ≈ 88% (approx.) S377. Ans.(e) Sol. Pattern is — +25, +50, +100, +200, +400 So, 402+400=802 S386. Ans.(d) Sol. I . x2 − 6𝑥 − 8𝑥 + 48 = 0 x(x −6) − 8 (𝑥 − 6) = 0 (x – 8) (𝑥 − 6) = 0 x = 6, 8 II . y2 – 9𝑦 − 8𝑦 + 72 = 0 y(y −9) − 8 (𝑦 − 9) = 0 (y −9) (𝑦 − 8) = 0 y = 9, 8 x ≤y S378. Ans.(a) Sol. Pattern is — +(5²–1), +(7²+1), +(9²–1), +(11²+1), +(13²–1) So, 293+(13²–1)=461 S379. Ans.(d) Sol. Pattern is — ÷5, ×6, ÷5, ×6, ÷5 So, 50.4 ÷ 5 = 10.08 S380. Ans.(b) Sol. ×2.5, ×1.5, ×2.5, ×1.5, ×2.5 So, 337.5 × 2.5 = 843.75 S381. Ans.(c) Sol. 19000 Marked price of C in 2018 = 100−24 × 100 = 25000. Cost price of C in 2018 = 25000 5 × 3 = 15000 Required Difference= 6000 − 4000 = 2000 S382. Ans.(d) Sol. Required average = 16000+25000+15000+22000 4 = 19500 𝑅𝑠. S383. Ans.(c) Sol. 19000 Required ratio = 25000 = 19: 25 88 www.bankersadda.com S387. Ans.(b) Sol. I . x2 + 7x + 6x + 42 = 0 𝑥 (𝑥 + 7) + 6(𝑥 + 7) = 0 (𝑥 + 6)(𝑥 + 7) = 0 x = −6, −7 II. y2 + 8y + 7y + 56 = 0 y(y + 8) + 7(y + 8) = 0 (y + 8) (y + 7) = 0 y = −8, −7 x≥y S388. Ans.(c) Sol. I . x2 + 6x + 2x+ 12 = 0 x(x + 6) + 2(x + 6) = 0 (x + 2) (x + 6) = 0 x = −2, −6 II. 6y2 + 9y + 4y + 6 = 0 3y(2y + 3) + 2(2y + 3) = 0 (2y + 3) (3y + 2) = 0 3 2 y =−2,−3 x<y | www.sscadda.com | www.careerpower.in | Adda247 App S389. Ans.(a) Sol. I . 2x2 + 6x + 3x + 9 = 0 2x(x + 3) + 3(x + 3) = 0 (x + 3) (2x + 3) = 0 3 x = −3, − 2 S395. Ans.(d) Sol. Pattern is, II . y2 + 16y + 12y + 192 = 0 y(y + 16) + 12(y + 16) = 0 (y + 16) (y + 12) = 0 y = −16, −12 x>y S396. Ans.(e) Sol. Price of a one kg sugar = 84 × S390. Ans.(a) Sol. I. x² – 9x + 20 = 0 x² – 5x – 4x + 20 = 0 x (x – 5) –4 (x – 5) = 0 (x – 4) (x – 5) = 0 x = 4, 5 II. 𝑦 2 + 3𝑦 + 3𝑦 + 9 = 0 𝑦(𝑦 + 3) + 3(𝑦 + 3) = 0 (𝑦 + 3)(𝑦 + 3) = 0 𝑦 = −3, −3 𝑥>𝑦 Price of one kg of salt = 840 × 10 21 11 21 = 𝑅𝑠 44 = 𝑅𝑠40 Required difference = (20 × 44 – 15 × 40) = 880 – 600 = 𝑅𝑠. 280 S397. Ans.(a) Sol. Price of one kg of tea = 900 18 = 𝑅𝑠50 1500 Price of one kg of rice = 30 = 𝑅𝑠50 Required % = 50−50 50 × 100 = 0% S398. Ans.(d) S391. Ans.(b) Sol. 63×12 18 Sol. Required ratio = 42×25 = 25 S399. Ans.(b) Required%= S392. Ans.(d) Sol. 20+15 30+12 1 × 100 = 83 % 3 S400. Ans.(b) Sol. Required sum = (56 × 15) + (32 × 30) + (40 × 25) = 2800 Rs. S393. Ans.(e) Sol. S394. Ans.(a) Sol. Pattern is, 89 www.bankersadda.com | www.sscadda.com | www.careerpower.in | Adda247 App
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